This blog post is intended to deliver a quick explanation of the algebra of Borel measures on $\mathbb{R}^n$. It will be broken into pieces. All complex-valued complex Borel measures $M(\mathbb{R}^n)$ clearly form a vector space over $\mathbb{C}$. The main goal of this post is to show that this is a Banach space and also a Banach algebra.

In fact, the $\mathbb{R}^n$ case can be generalised into any locally compact abelian group (see any abstract harmonic analysis books), this is because what really matters here is being locally compact and abelian. But at this moment we stick to Euclidean spaces. Note since $\mathbb{R}^n$ is $\sigma$-compact, all Borel measures are regular.

To read this post you need to be familiar with some basic properties of Banach algebra, complex Borel measures, and the most important, Fubini’s theorem.

Suppose $1 < p < \infty$ and $f \in L^p((0,\infty))$ (with respect to Lebesgue measure of course) is a nonnegative function, take

$$F(x) = \frac{1}{x}\int_0^x f(t)dt \quad 0 < x <\infty,$$

we have Hardy’s inequality $\def\lrVert[#1]{\lVert #1 \rVert}$

$$\lrVert[F]_p \leq q\lrVert[f]_p$$

where $\frac{1}{p}+\frac{1}{q}=1$ of course.

There are several ways to prove it. I think there are several good reasons to write them down thoroughly since that may be why you find this page. Maybe you are burnt out since it’s left as exercise. You are assumed to have enough knowledge of Lebesgue measure and integration.

Minkowski’s integral inequality

Let $S_1,S_2 \subset \mathbb{R}$ be two measurable set, suppose $F:S_1 \times S_2 \to \mathbb{R}$ is measurable, then

$$\left[\int_{S_2} \left\vert\int_{S_1}F(x,y)dx \right\vert^pdy\right]^{\frac{1}{p}} \leq \int_{S_1} \left[\int_{S_2} |F(x,y)|^p dy\right]^{\frac{1}{p}}dx.$$

A proof can be found at here by turning to Example A9. You may need to replace all measures with Lebesgue measure $m$.

Now let’s get into it. For a measurable function in this place we should have $G(x,t)=\frac{f(t)}{x}$. If we put this function inside this inequality, we see

$$\begin{aligned} \lrVert[F]_p &= \left[\int_0^\infty \left\vert \int_0^x \frac{f(t)}{x}dt \right\vert^p dx\right]^{\frac{1}{p}} \\ &= \left[\int_0^\infty \left\vert \int_0^1 f(ux)du \right\vert^p dx\right]^{\frac{1}{p}} \\ &\leq \int_0^1 \left[\int_0^\infty |f(ux)|^pdx\right]^{\frac{1}{p}}du \\ &= \int_0^1 \left[\int_0^\infty |f(ux)|^pudx\right]^{\frac{1}{p}}u^{-\frac{1}{p}}du \\ &= \lrVert[f]_p \int_0^1 u^{-\frac{1}{p}}du \\ &=q\lrVert[f]_p. \end{aligned}$$

Note we have used change-of-variable twice and the inequality once.

A constructive approach

I have no idea how people came up with this solution. Take $xF(x)=\int_0^x f(t)t^{u}t^{-u}dt$ where $0<u<1-\frac{1}{p}$. Hölder’s inequality gives us

$$\begin{aligned} xF(x) &= \int_0^x f(t)t^ut^{-u}dt \\ &\leq \left[\int_0^x t^{-uq}dt\right]^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \\ &=\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}} \end{aligned}$$

Hence

$$\begin{aligned} F(x)^p & \leq \frac{1}{x^p}\left\{\left(\frac{1}{1-uq}x^{1-uq}\right)^{\frac{1}{q}}\left[\int_0^xf(t)^pt^{up}dt\right]^{\frac{1}{p}}\right\}^{p} \\ &= \left(\frac{1}{1-uq}\right)^{\frac{p}{q}}x^{\frac{p}{q}(1-uq)-p}\int_0^x f(t)^pt^{up}dt \\ &= \left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt \end{aligned}$$

Note we have used the fact that $\frac{1}{p}+\frac{1}{q}=1 \implies p+q=pq$ and $\frac{p}{q}=p-1$. Fubini’s theorem gives us the final answer:

$$\begin{aligned} \int_0^\infty F(x)^pdx &\leq \int_0^\infty\left[\left(\frac{1}{1-uq}\right)^{p-1}x^{-up-1}\int_0^x f(t)^pt^{up}dt\right]dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dx\int_0^x f(t)^pt^{up}x^{-up-1}dt \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\int_0^\infty dt\int_t^\infty f(t)^pt^{up}x^{-up-1}dx \\ &=\left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}\int_0^\infty f(t)^pdt. \end{aligned}$$

It remains to find the minimum of $\varphi(u) = \left(\frac{1}{1-uq}\right)^{p-1}\frac{1}{up}$. This is an elementary calculus problem. By taking its derivative, we see when $u=\frac{1}{pq}<1-\frac{1}{p}$ it attains its minimum $\left(\frac{p}{p-1}\right)^p=q^p$. Hence we get

$$\int_0^\infty F(x)^pdx \leq q^p\int_0^\infty f(t)^pdt,$$

which is exactly what we want. Note the constant $q$ cannot be replaced with a smaller one. We simply proved the case when $f \geq 0$. For the general case, one simply needs to take absolute value.

Integration by parts

This approach makes use of properties of $L^p$ space. Still we assume that $f \geq 0$ but we also assume $f \in C_c((0,\infty))$, that is, $f$ is continuous and has compact support. Hence $F$ is differentiable in this situation. Integration by parts gives

$$\int_0^\infty F^p(x)dx=xF(x)^p\vert_0^\infty- p\int_0^\infty xdF^p = -p\int_0^\infty xF^{p-1}(x)F'(x)dx.$$

Note since $f$ has compact support, there are some $[a,b]$ such that $f >0$ only if $0 < a \leq x \leq b < \infty$ and hence $xF(x)^p\vert_0^\infty=0$. Next it is natural to take a look at $F’(x)$. Note we have

$$F'(x) = \frac{f(x)}{x}-\frac{\int_0^x f(t)dt}{x^2},$$

hence $xF’(x)=f(x)-F(x)$. A substitution gives us

$$\int_0^\infty F^p(x)dx = -p\int_0^\infty F^{p-1}(x)[f(x)-F(x)]dx,$$

which is equivalent to say

$$\int_0^\infty F^p(x)dx = \frac{p}{p-1}\int_0^\infty F^{p-1}(x)f(x)dx.$$

Hölder’s inequality gives us

$$\begin{aligned} \int_0^\infty F^{p-1}(x)f(x)dx &\leq \left[\int_0^\infty F^{(p-1)q}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}} \\ &=\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}. \end{aligned}$$

Together with the identity above we get

$$\int_0^\infty F^p(x)dx = q\left[\int_0^\infty F^{p}(x)dx\right]^{\frac{1}{q}}\left[\int_0^\infty f(x)^pdx\right]^{\frac{1}{p}}$$

which is exactly what we want since $1-\frac{1}{q}=\frac{1}{p}$ and all we need to do is divide $\left[\int_0^\infty F^pdx\right]^{1/q}$ on both sides. So what’s next? Note $C_c((0,\infty))$ is dense in $L^p((0,\infty))$. For any $f \in L^p((0,\infty))$, we can take a sequence of functions $f_n \in C_c((0,\infty))$ such that $f_n \to f$ with respect to $L^p$-norm. Taking $F=\frac{1}{x}\int_0^x f(t)dt$ and $F_n = \frac{1}{x}\int_0^x f_n(t)dt$, we need to show that $F_n \to F$ pointwise, so that we can use Fatou’s lemma. For $\varepsilon>0$, there exists some $m$ such that $\lrVert[f_n-f]_p < \frac{1}{n}$. Thus

$$\begin{aligned} |F_n(x)-F(x)| &= \frac{1}{x}\left\vert \int_0^x f_n(t)dt - \int_0^x f(t)dt \right\vert \\ &\leq \frac{1}{x} \int_0^x |f_n(t)-f(t)|dt \\ &\leq \frac{1}{x} \left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}}\left[\int_0^x 1^qdt\right]^{\frac{1}{q}} \\ &=\frac{1}{x^{1/p}}\left[\int_0^x|f_n(t)-f(t)|^pdt\right]^{\frac{1}{p}} \\ &\leq \frac{1}{x^{1/p}}\lrVert[f_n-f]_p <\frac{\varepsilon}{x^{1/p}}. \end{aligned}$$

Hence $F_n \to F$ pointwise, which also implies that $|F_n|^p \to |F|^p$ pointwise. For $|F_n|$ we have

$$\begin{aligned} \int_0^\infty |F_n(x)|^pdx &= \int_0^\infty \left\vert\frac{1}{x}\int_0^x f_n(t)dt\right\vert^p dx \\ &\leq \int_0^\infty \left[\frac{1}{x}\int_0^x |f_n(t)|dt\right]^{p}dx \\ &\leq q\int_0^\infty |f_n(t)|^pdt \end{aligned}$$

note the third inequality follows since we have already proved it for $f \geq 0$. By Fatou’s lemma, we have

$$\begin{aligned} \int_0^\infty |F(x)|^pdx &= \int_0^\infty \lim_{n \to \infty}|F_n(x)|^pdx \\ &\leq \lim_{n \to \infty} \int_0^\infty |F_n(x)|^pdx \\ &\leq \lim_{n \to \infty}q^p\int_0^\infty |f_n(x)|^pdx \\ &=q^p\int_0^\infty |f(x)|^pdx. \end{aligned}$$

Throughout, let $(X,\mathfrak{M},\mu)$ be a measure space where $\mu$ is positive.

The question

If $f$ is of $L^p(\mu)$, which means $\lVert f \rVert_p=\left(\int_X |f|^p d\mu\right)^{1/p}<\infty$, or equivalently $\int_X |f|^p d\mu<\infty$, then we may say $|f|^p$ is of $L^1(\mu)$. In other words, we have a function

$$\begin{aligned} \lambda: L^p(\mu) &\to L^1(\mu) \\ f &\mapsto |f|^p. \end{aligned}$$

This function does not have to be one to one due to absolute value. But we hope this function to be fine enough, at the very least, we hope it is continuous.

Here, $f \sim g$ means that $f-g$ equals $0$ almost everywhere with respect to $\mu$. It can be easily verified that this is an equivalence relation.

Continuity

We still use the $\varepsilon-\delta$ argument but it’s in a metric space. Suppose $(X,d_1)$ and $(Y,d_2)$ are two metric spaces and $f:X \to Y$ is a function. We say $f$ is continuous at $x_0 \in X$ if, for any $\varepsilon>0$, there exists some $\delta>0$ such that $d_2(f(x_0),f(x))<\varepsilon$ whenever $d_1(x_0,x)<\delta$. Further, we say $f$ is continuous on $X$ if $f$ is continuous at every point $x \in X$.

Metrics

For $1\leq p<\infty$, we already have a metric by

$$d(f,g)=\lVert f-g \rVert_p$$

given that $d(f,g)=0$ if and only if $f \sim g$. This is complete and makes $L^p$ a Banach space. But for $0<p<1$ (yes we are going to cover that), things are much more different, and there is one reason: Minkowski inequality holds reversely! In fact, we have

$$\lVert f+g \rVert_p \geq \lVert f \rVert_p + \lVert g \rVert_p$$

for $0<p<1$. $L^p$ space has too many weird things when $0<p<1$. Precisely,

For $0<p<1$, $L^p(\mu)$ is locally convex if and only if $\mu$ assumes finitely many values. (Proof.)

On the other hand, for example, $X=[0,1]$ and $\mu=m$ be the Lebesgue measure, then $L^p(\mu)$ has no open convex subset other than $\varnothing$ and $L^p(\mu)$ itself. However,

A topological vector space $X$ is normable if and only if its origin has a convex bounded neighbourhood. (See Kolmogorov’s normability criterion.)

Therefore $L^p(m)$ is not normable, hence not Banach.

We have gone too far. We need a metric that is fine enough.

Metric of $L^p$ when $0<p<1$

Define

$$\Delta(f)=\int_X |f|^p d\mu$$

for $f \in L^p(\mu)$. We will show that we have a metric by

$$d(f,g)=\Delta(f-g).$$

Fix $y\geq 0$, consider the function

$$f(x)=(x+y)^p-x^p.$$

We have $f(0)=y^p$ and

$$f'(x)=p(x+y)^{p-1}-px^{p-1} \leq px^{p-1}-px^{p-1}=0$$

when $x > 0$ and hence $f(x)$ is nonincreasing on $[0,\infty)$, which implies that

$$(x+y)^p \leq x^p+y^p.$$

Hence for any $f$, $g \in L^p$, we have

$$\Delta(f+g)=\int_X |f+g|^p d\mu \leq \int_X |f|^p d\mu + \int_X |g|^p d\mu=\Delta(f)+\Delta(g).$$

This inequality ensures that

$$d(f,g)=\Delta(f-g)$$

is a metric. It’s immediate that $d(f,g)=d(g,f) \geq 0$ for all $f$, $g \in L^p(\mu)$. For the triangle inequality, note that

$$d(f,h)+d(g,h)=\Delta(f-h)+\Delta(h-g) \geq \Delta((f-h)+(h-g))=\Delta(f-g)=d(f,g).$$

This is translate-invariant as well since

$$d(f+h,g+h)=\Delta(f+h-g-h)=\Delta(f-g)=d(f,g)$$

The completeness can be verified in the same way as the case when $p>1$. In fact, this metric makes $L^p$ a locally bounded F-space.

The continuity of $\lambda$

The metric of $L^1$ is defined by

$$d_1(f,g)=\lVert f-g \rVert_1=\int_X |f-g|d\mu.$$

We need to find a relation between $d_p(f,g)$ and $d_1(\lambda(f),\lambda(g))$, where $d_p$ is the metric of the corresponding $L^p$ space.

$0<p<1$

As we have proved,

$$(x+y)^p \leq x^p+y^p.$$

Without loss of generality we assume $x \geq y$ and therefore

$$x^p=(x-y+y)^p \leq (x-y)^p+y^p.$$

Hence

$$x^p-y^p \leq (x-y)^p.$$

By interchanging $x$ and $y$, we get

$$|x^p-y^p| \leq |x-y|^p.$$

Replacing $x$ and $y$ with $|f|$ and $|g|$ where $f$, $g \in L^p$, we get

$$\int_{X}\lvert |f|^p-|g|^p \rvert d\mu \leq \int_X |f-g|^p d\mu.$$

But

$$d_1(\lambda(f),\lambda(g))=\int_{X}\lvert |f|^p-|g|^p \rvert d\mu \\ d_p(f,g)=\Delta(f-g)= d\mu \leq \int_X |f-g|^p d\mu$$

and we therefore have

$$d_1(\lambda(f),\lambda(g)) \leq d_p(f,g).$$

Hence $\lambda$ is continuous (and in fact, Lipschitz continuous and uniformly continuous) when $0<p<1$.

$1 \leq p < \infty$

It’s natural to think about Minkowski’s inequality and Hölder’s inequality in this case since they are critical inequality enablers. You need to think about some examples of how to create the condition to use them and get a fine result. In this section we need to prove that

$$|x^p-y^p| \leq p|x-y|(x^{p-1}+y^{p-1}).$$

This inequality is surprisingly easy to prove however. We will use nothing but the mean value theorem. Without loss of generality we assume that $x > y \geq 0$ and define $f(t)=t^p$. Then

$$\frac{f(x)-f(y)}{x-y}=f'(\zeta)=p\zeta^{p-1}$$

where $y < \zeta < x$. But since $p-1 \geq 0$, we see $\zeta^{p-1} < x^{p-1} <x^{p-1}+y^{p-1}$. Therefore

$$f(x)-f(y)=x^p-y^p=p(x-y)\zeta^{p-1}<p(x-y)(x^{p-1}-y^{p-1}).$$

For $x=y$ the equality holds.

---

Therefore

$$\begin{aligned} d_1(\lambda(f),\lambda(g)) &= \int_X \left||f|^p-|g|^p\right|d\mu \\ &\leq \int_Xp\left||f|-|g|\right|(|f|^{p-1}+|g|^{p-1})d\mu \end{aligned}$$

By Hölder’s inequality, we have

$$\begin{aligned} \int_X ||f|-|g||(|f|^{p-1}+|g|^{p-1})d\mu & \leq \left[\int_X \left||f|-|g|\right|^pd\mu\right]^{1/p}\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \\ &\leq \left[\int_X \left|f-g\right|^pd\mu\right]^{1/p}\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \\ &=\lVert f-g \rVert_p \left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q}. \end{aligned}$$

By Minkowski’s inequality, we have

$$\left[\int_X\left(|f|^{p-1}+|g|^{p-1}\right)^q\right]^{1/q} \leq \left[\int_X|f|^{(p-1)q}d\mu\right]^{1/q}+\left[\int_X |g|^{(p-1)q}d\mu\right]^{1/q}$$

Now things are clear. Since $1/p+1/q=1$, or equivalently $1/q=(p-1)/p$, suppose $\lVert f \rVert_p$, $\lVert g \rVert_p \leq R$, then $(p-1)q=p$ and therefore

$$\left[\int_X|f|^{(p-1)q}d\mu\right]^{1/q}+\left[\int_X |g|^{(p-1)q}d\mu\right]^{1/q} = \lVert f \rVert_p^{p-1}+\lVert g \rVert_p^{p-1} \leq 2R^{p-1}.$$

Summing the inequalities above, we get

$$\begin{aligned} d_1(\lambda(f),\lambda(g)) \leq 2pR^{p-1}\lVert f-g \rVert_p =2pR^{p-1}d_p(f,g) \end{aligned}$$

hence $\lambda$ is continuous.

Conclusion and further

We have proved that $\lambda$ is continuous, and when $0<p<1$, we have seen that $\lambda$ is Lipschitz continuous. It’s natural to think about its differentiability afterwards, but the absolute value function is not even differentiable so we may have no chance. But this is still a fine enough result. For example we have no restriction to $(X,\mathfrak{M},\mu)$ other than the positivity of $\mu$. Therefore we may take $\mathbb{R}^n$ as the Lebesgue measure space here, or we can take something else.

It’s also interesting how we use elementary Calculus to solve some much more abstract problems.

In this post we develop a proof of the Riesz-Markov-Kakutani theorem on a locally compact Hausdorff space, which is the essential of the existence of the Lebesgue measure.

An introduction

If one wants to learn the fundamental theorem of Calculus in the sense of Lebesgue integral, properties of measures have to be taken into account. In elementary calculus, one may consider something like

$$df(x)=f'(x)dx$$

where $f$ is differentiable, say, everywhere on an interval. Now we restrict $f$ to be a differentiable and nondecreasing real function defined on $I=[a,b]$. There we got a one-to-one function defined by

$$g(x)=x+f(x)$$

For measurable sets $E\in\mathfrak{M}$, it can be seen that if $m(E)=0$, we have $m(g(E))=0$. Moreover, $g(E) \in \mathfrak{M}$, and $g$ is one-to-one. Therefore we can define a measure like

$$\mu(E)=m(g(E))$$

If we have a relation

$$\mu(E)=\int_{E}hdm$$

(in fact, this is the Radon-Nikodym theorem we will prove later), the fundamental theorem of calculus for $f$ becomes somewhat clear since if $E=[a,x]$, we got $g(E)=[a+f(a),x+f(x)]$, thus we got

$$\begin{aligned} \mu(E)=m(g(E))&=g(x)-g(a)\\ &=f(x)-f(a)+\int_a^xdt \\ &=\int_a^xh(t)dt \end{aligned}$$

which trivially implies

$$f(x)-f(a)=\int_a^x[h(t)-1]dt$$

the function $h$ looks like to be $g’=f’+1$.

We are not proving the fundamental theorem here. But this gives rise to a question. Is it possible to find a function such that

$$\mu(E)=\int_{E}hdm$$

one may write as

$$d\mu=hdm$$

or, more generally, a measure $\mu$ with respect to another measure $\lambda$? Does this $\mu$ exist with respect to $\lambda$? Does this $h$ exist? Lot of questions. Luckily the Lebesgue decomposition and Radon-Nikodym theorem make it possible.

Notations

Let $\mu$ be a positive measure on a $\sigma$-algebra $\mathfrak{M}$, let $\lambda$ be any arbitrary measure (positive or complex) defined on $\mathfrak{M}$.

We write

$$\lambda \ll \mu$$

if $\lambda(E)=0$ for every $E\in\mathfrak{M}$ for which $\mu(E)=0$. (You may write $\mu \ll m$ in the previous section.) We say $\lambda$ is absolutely continuous with respect to $\mu$.

Another relation between measures worth consideration is being mutually singular. If we have $\lambda(E)=\lambda(A \cap E)$ for every $E \in \mathfrak{M}$, we say $\lambda$ is concentrated on $A$.

If we now have two measures $\mu_1$ and $\mu_2$, two disjoint sets $A$ and $B$ such that $\mu_1$ is concentrated on $A$, $\mu_2$ is concentrated on $B$, we say $\mu_1$ and $\mu_2$ are mutually singular, and write

$$\mu_1 \perp \mu_2$$

The Theorem of Lebesgue-Radon-Nikodym

Let $\mu$ be a positive $\sigma$-finite measure on $\mathfrak{M}$, and $\lambda$ a complex measure on $\mathfrak{M}$.

• There exists a unique pair of complex measures $\lambda_{ac}$ and $\lambda_{s}$ on $\mathfrak{M}$ such that

$$\lambda = \lambda_{ac}+\lambda_s \quad \lambda_{ac}\ll\mu\quad \lambda_s \perp \mu$$

• There is a unique $h \in L^1(\mu)$ such that

$$\lambda_{ac}(E)=\int_{E}hd\mu$$

for every $E \in \mathfrak{M}$.

The unique pair $(\lambda_{ac},\lambda_s)$ is called the Lebesgue decomposition; the existence of $h$ is called the Radon-Nikodym theorem, and $h$ is called the Radon-Nikodym derivative. One also writes $d\lambda_{ac}=hd\mu$ or $\frac{d\lambda_{ac}}{d\mu}=h$ in this situation.

These are two separate theorems, but von Neumann gave the idea to prove these two at one stroke.

If we already have $\lambda \ll \mu$, then $\lambda_s=0$ and the Radon-Nikodym derivative shows up in the natural of things.

Also, one cannot ignore the fact that $m$ the Lebesgue measure is $\sigma$-finite.

Proof explained

Step 1 - Construct a bounded functional

We are going to employ Hilbert space technique in this proof. Precisely speaking, we are going to construct a bounded linear functional to find another function, namely $g$, which is the epicentre of this proof.

The boundedness of $\lambda$ is clear since it’s complex, but $\mu$ is only assumed to be $\sigma$-finite. Therefore we need some adjustment onto $\mu$.

1.1 Replacing $\mu$ with a finite measure

If $\mu$ is a positive $\sigma$-finite measure on a $\sigma$-algebra $\mathfrak{M}$ in a set $X$, then there is a function $w$ such that $w \in L^1(\mu)$ and $0<w(x)<1$ for every $x \in X$.

The $\sigma$-finiteness of $\mu$ denotes that, there exist some sets $E_n$ such that

$$X=\bigcup_{n=1}^{\infty}E_n$$

and that $\mu(E_n)<\infty$ for all $n$.

Define

$$w_n(x)= \begin{aligned} \begin{cases} \frac{1}{2^n(1+\mu(E_n))}\quad &x \in E_n \\ 0 \quad &x\notin E_n \end{cases} \end{aligned}$$

(you can also say that $w_n=\frac{1}{2^n(1+\mu(E_n))}\chi_{E_n}$), then we have

$$\begin{aligned} w &= \sum_{n=1}^{\infty}w_n \\ \end{aligned}$$

satisfies $0<w<1$ for all $x$. With $w$, we are able to define a new measure, namely

$$\tilde{\mu}(E)=\int_{E}wd\mu.$$

The fact that $\tilde{\mu}(E)$ is a measure can be validated by considering $\int_{E}wd\mu=\int_{X}\chi_{E}wd\mu$. It’s more important that $\tilde{\mu}(E)$ is bounded and $\tilde{\mu}(E)=0$ if and only if $\mu(E)=0$. The second one comes from the strict positivity of $w$. For the first one, notice that

$$\begin{aligned} \tilde{\mu}(X) &\leq \sum_{n=1}^{\infty}\tilde{\mu}(E_n) \\ &= \sum_{n=1}^{\infty}\frac{1}{2^n(1+\mu(E_n))} \\ &\leq \sum_{n=1}^{\infty}\frac{1}{2^n} \end{aligned}$$

1.2 A bounded linear functional associated with $\lambda$

Since $\lambda$ is complex, without loss of generality, we are able to assume that $\lambda$ is a positive bounded measure on $\mathfrak{M}$. By 1.1, we are able to obtain a positive bounded measure by

$$\varphi=\lambda+\tilde{\mu}$$

Following the construction of Lebesgue measure, we have

$$\int_{X}fd\varphi=\int_{X}fd\lambda+\int_{X}fwd\mu$$

for all nonnegative measurable function $f$. Also, notice that $\lambda \leq \varphi$, we have

$$\left\vert \int_{X}fd\lambda \right\vert \leq \int_{X}|f|d\lambda \leq \int_{X}|f|d\varphi \leq \sqrt{\varphi(X)}\left\Vert f \right\Vert_2$$

for $f \in L^2(\varphi)$ by Schwarz inequality.

Since $\varphi(X)<\infty$, we have

$$\Lambda{f}=\int_{X}fd\lambda$$

to be a bounded linear functional on $L^2(\varphi)$.

Step 2 - Find the associated function with respect to $\lambda$

Since $L^2(\varphi)$ is a Hilbert space, every bounded linear functional on a Hilbert space $H$ is given by an inner product with an element in $H$. That is, by the completeness of $L^2(\varphi)$, there exists a function $g$ such that

$$\Lambda{f}=\int_{X}fd\lambda=\int_{X}fgd\varphi=(f,g).$$

The properties of $L^2$ space shows that $g$ is determined almost everywhere with respect to $\varphi$.

For $E \in \mathfrak{M}$, we got

$$0 \leq (\chi_{E},g)=\int_{E}gd\varphi=\int_{E}d\lambda=\lambda(E)\leq\varphi(E)$$

which implies $0 \leq g \leq 1$ for almost every $x$ with respect to $\varphi$. Therefore we are able to assume that $0 \leq g \leq 1$ without ruining the identity. The proof is in the bag once we define $A$ to be the set where $0 \leq g < 1$ and $B$ the set where $g=1$.

Step 3 - Generate $\lambda_{ac}$ and $\lambda_{s}$ and the Radon-Nikodym derivative at one stroke

We claim that $\lambda(A \cap E)$ and $\lambda(B \cap E)$ form the decomposition we are looking for, $\lambda_{ac}$ and $\lambda_s$, respectively. Namely, $\lambda_{ac}=\lambda(A \cap E)$, $\lambda_s=\lambda(B \cap E)$.

Proving $\lambda_s \perp \mu$

If we combine $\Lambda{f}=(f,g)$ and $\varphi=\lambda+\tilde{\mu}$ together, we have

$$\int_{X}(1-g)fd\lambda=\int_{X}fgwd\mu.$$

Put $f=\chi_{B}$, we have

$$\int_{B}wd\mu=0.$$

Since $w$ is strictly positive, we see that $\mu(B)=0$. Notice that $A \cap B = \varnothing$ and $A \cup B=X$. For $E \in \mathfrak{M}$, we write $E=E_A \cup E_B$, where $E_A \subset A$ and $E_B \subset B$. Therefore

$$\mu(E)=\mu(E_A)+\mu(E_B)=\mu(E \cap A)+\mu(E \cap B)=\mu(E \cap A).$$

Therefore $\mu$ is concentrated on $A$.

For $\lambda_s$, observe that

$$\lambda_s(E)=\lambda(E \cap B)=\lambda((E \cap B) \cap B)=\lambda_s(E \cap B).$$

Hence $\lambda_s$ is concentrated on $B$. This observation shows that $\lambda_s \perp \mu$.

Proving $\lambda_{ac} \ll \mu$ by the Radon-Nikodym derivative

The relation that $\lambda_{ac} \ll \mu$ will be showed by the existence of the Radon-Nikodym derivative.

If we replace $f$ by

$$(1+g+\cdots+g^n)\chi_E,$$

where $E \in \mathfrak{M}$, we have

$$\int_X(1-g)fd\lambda=\int_E(1-g^{n+1})d\lambda=\int_Eg(1+g+\cdots+g^n)wd\mu.$$

Notice that

$$\begin{aligned} \int_{E}(1-g^{n+1})d\lambda &=\int\limits_{E \cap A}(1-g^{n+1})d\lambda + \int\limits_{E \cap B}(1-g^{n+1})d\lambda \\ &=\int\limits_{E \cap A}(1-g^{n+1})d\lambda \\ &\to\lambda(E \cap A) = \lambda_{ac}(E)\quad(n\to\infty) \end{aligned}$$

Define $h_n=g(1+g+g^2+\cdots+g^n)w$, we see that on $A$, $h_n$ converges monotonically to

$$h= \begin{aligned} \begin{cases} \frac{gw}{1-g} \quad &x\in{A}\\ 0 \quad &x\in{B} \end{cases} \end{aligned}$$

By monotone convergence theorem, we got

$$\lim_{n\to\infty}\int_{E}h_nd\mu = \int_{E}hd\mu=\lambda_{ac}(E).$$

for every $E\in\mathfrak{M}$.

The measurable function $h$ is the desired Radon-Nikodym derivative once we show that $h \in L^1(\mu)$. Replacing $E$ with $X$, we see that

$$\int_{X}|h|d\mu=\int_{X}hd\mu=\lambda_{ac}(X)\leq\lambda(X)<\infty.$$

Clearly, if $\mu(E)=0$, we have

$$\lambda_{ac}(E)=\int_{E}hd\mu=0$$

which shows that

$$\lambda_{ac}\ll\mu$$

as desired.

Step 3 - Generalization onto complex measures

By far we have proved this theorem for positive bounded measure. For real bounded measure, we can apply the proceeding case to the positive and negative part of it. For all complex measures, we have

$$\lambda=\lambda_1+i\lambda_2$$

where $\lambda_1$ and $\lambda_2$ are real.

Step 4 - Uniqueness of the decomposition

If we have two Lebesgue decompositions of the same measure, namely $(\lambda_{ac},\lambda_s)$ and $(\lambda’_{ac},\lambda’_s)$, we shall show that

$$\lambda_{ac}-\lambda_{ac}'=\lambda_s'-\lambda_s=0$$

By the definition of the decomposition we got

$$\lambda_{ac}-\lambda'_{ac}=\lambda'_s-\lambda_s$$

with $\lambda_{ac}-\lambda_{ac}’ \ll \mu$ and $\lambda_{s}’-\lambda_{s}\perp\mu$. This implies that $\lambda’_{s}-\lambda_{s} \ll \mu$ as well.

Since $\lambda’_s-\lambda_s\perp\mu$, there exists a set with $\mu(A)=0$ on which $\lambda’_s-\lambda_s$ is concentrated; the absolute continuity shows that $\lambda’_s(E)-\lambda_s(E)=0$ for all $E \subset A$. Hence $\lambda_s’-\lambda_s$ is concentrated on $X-A$. Therefore we got $(\lambda’_s-\lambda_s)\perp(\lambda’_s-\lambda_s)$, which forces $\lambda’_s-\lambda_s=0$. The uniqueness is proved.

(Following the same process one can also show that $\lambda_{ac}\perp\lambda_s$.)