Integration

The $(\mu)$-integral of a non-negative simple function is by definition

$$\int_Y f d \mu = \sum_i a_i \mu (Y \cap E_i) , Y \in \mathcal{M} .$$ (26)

Here the convention is that if $\mu (Y \cap E_i) = \infty$ but $a_i=0$ then $a_i \cdot \mu (Y \cap E_i) =0$. Clearly this integral takes values in $[0,\infty ]$. More significantly, if $c\ge0$ is a constant and $f$ and $g$ are two non-negative ($\mu$-measurable) simple functions then

$$\begin{gathered}\int_Y cfd\mu=c\int_Y fd\mu\ \int_Y(f+g)d\mu... ... \int_Y f d \mu \leq \int_Y g d \mu . \end{gathered}$$ (27)

(See [1] Proposition 2.13 on page 48.)

To see this, observe that (3.1) holds for any presentation (2.14) of $f$ with all $a_i \geq 0$. Indeed, by restriction to $E_i$ and division by $a_i$ (which can be assumed non-zero) it is enough to consider the special case

$$\chi_E=\sum\limits_{j}b_j\chi_{F_j}.$$

The $F_j$ can always be written as the union of a finite number, $N',$ of disjoint measurable sets, $F_j=\cup_{l\in S_j}G_l$ where $j=1,\dots,N$ and $S_j\subset\{1,\dots,N'\}.$ Thus

$$\sum\limits_jb_j\mu(F_j)=\sum\limits_jb_j\sum\limits_{l\in S_j}\mu(G_l)=\mu(E)$$

since $\sum_{\{j;l\in S_j\}}b_j=1$ for each $j.$

From this all the statements follow easily.

By taking suprema, $\int_Efd\mu$ has the first and last properties in (3.2). It also has the middle property, but this is less obvious. To see this, we shall prove the basic `Monotone convergence theorem' (of Lebesgue). Before doing so however, note what the vanishing of the integral means.

Proof. If (3.4) holds, then any positive simple function bounded above by $f$ must also vanish outside a set of measure zero, so its integral must be zero and hence $\int_Efd\mu=0.$ Conversely, observe that the set in (3.4) can be written as

$$E_n=\bigcup_{n}\{x\in E;f(x)>1/n\}.$$

Since these increase, if (3.4) does not hold then one of these must have positive measure. In that case the simple function $n^{-1}\chi_{E_n}$ has positve integral so $\int_Efd\mu>0.$ $\qedsymbol$

Notice the fundamental difference in approach here between Riemann and Lebesgue integrals. The Lebesgue integral, (3.3), uses approximation by functions constant on possibly quite nasty measurable sets, not just intervals as in the Riemann lower and upper integrals.

Proof. To see that $f$ is measurable, observe that

$$f^{-1}(a,\infty]=\bigcup_nf_n^{-1}(a,\infty].$$ (28)

Since the sets $(a, \infty ]$ generate the Borel $\sigma$-algebra this shows that $f$ is measurable.

So we proceed to prove the main part of the proposition, which is (3.5). Rudin has quite a nice proof of this, [5] page 21. Here I paraphrase it. We can easily see from (3.1) that

$$\alpha=\sup\int_Ef_nd\mu=\lim_{n\to\infty}\int_Ef_nd\mu\le \int_E fd\mu.$$

Given a simple measurable function $g$ with $0\le g\le f$ and $0<c<1$ consider the sets $E_n=\{x\in E;f_n(x)\ge cg(x)\}.$ These are measurable and increase with $n.$ Moreover $E=\bigcup_nE_n.$ It follows that

$$\int_Ef_nd\mu\ge\int_{E_n}f_nd\mu\ge c\int_{E_n}gd\mu=\sum_ia_i\mu(E_n\cap F_i)$$ (29)

in terms of the natural presentation of $g=\sum_ia_i\chi_{F_i}.$ Now, the fact that the $E_n$ are measurable and increase to $E$ shows that

$$\mu(E_n\cap F_i)\to\mu(E\cap F_i)$$

as $n\to\infty.$ Thus the right side of (3.7) tends to $c\int_E gd\mu$ as $n\to\infty.$ Hence $\alpha\ge c\int_Egd\mu$ for all $0<c<1.$ Taking the supremum over $c$ and then over all such $g$ shows that

$$\alpha =\lim\limits_{n\to\infty}\int_Ef_n d\mu\ge \sup\int_Egd\mu=\int_Efd\mu.$$

They must therefore be equal. $\qedsymbol$

Now for instance the additivity in (3.1) for $f\ge0$ and $g\ge0$ any measurable functions follows from

Proof. Folland [1] page 45 has a nice proof. For each integer $n>0$ and $0\le k\le 2^{2n}-1,$ set

$$E_{n,k}=\{x\in X;2^{-n}k\le f(x)< 2^{-n}(k+1)\},$$

$$E_{n}'=\{x\in X;f(x)\ge2^{n}\}.$$

These are measurable sets. On increasing $n$ by one, the interval in the definition of $E_{n,k}$ is divided into two. It follows that the sequence of simple functions

$$f_n=\sum\limits_{k} 2^{-n}k\chi_{E_{k,n}}+2^{n}\chi_{E'_n}$$ (30)

is increasing and has limit $f$ and that this limit is uniform on any measurable set where $f$ is finite. $\qedsymbol$

Thus $\int_Efd\mu=\lim_{n\to\infty}\int_Ef_nd\mu$ and if $f$ and $g$ are two non-negative measurable functions then $f_n(x)+g_n(x)\uparrow f+g(x)$ so indeed

$$\int_E(f+g)d\mu=\int_Efd\mu+\int_Egd\mu.$$

As with the definition of $u_+$ long ago, this allows us to extend the definition of the integral to any integrable function.

Notice if $f$ is $\mu$-integrable then so is $\vert f\vert.$ One of the objects we wish to study is the space of integrable functions. The fact that the integral of $\vert f\vert$ can vanish encourages us to look at what at first seems a much more complicated object. Namely we consider an equivalence relation between integrable functions

$$f_1\equiv f_2\Longleftrightarrow \mu(\{x\in X;f_1(x)\not=f_2(x)\})=0.$$ (31)

That is we identify two such functions if they are equal `off a set of measure zero.' Clearly if $f_1\equiv f_2$ in this sense then

$$\int_X\vert f_1\vert d\mu=\int_X\vert f_2\vert d\mu=0, \int_Xf_1d\mu=\int_Xf_2d\mu.$$

A necessary condition for a measurable function $f\ge0$ to be integrable is

$$\mu\{x\in X;f(x)=\infty\}=0.$$

Let $E$ be the (necessarily measureable) set where $f=\infty.$ Indeed, if this does not have measure zero, then the sequence of simple functions $n\chi_E\le f$ has integral tending to infinity. It follows that each equivalence class under (3.9) has a representative which is an honest function, i.e. which is finite everywhere. Namely if $f$ is one representative then

$$f'(x)=\begin{cases}f(x) &x\notin E\ 0&x\in E\end{cases}$$

is also a representative.

We shall denote by $L^1(X,\mu)$ the space consisting of such equivalence classes of integrable functions. This is a normed linear space as I ask you to show in Problem 11.

The monotone convergence theorem often occurrs in the slightly disguised form of Fatou's Lemma.

Proof. Set $F_k (x) = \inf_{n \geq k} f_n (x)$. Thus $F_k$ is an increasing sequence of non-negative functions with limiting function $\liminf_{n\to\infty} f_n$ and $F_k (x) \leq f_n (x) \forall n \geq k$. By the monotone convergence theorem

$$\int \liminf_{n \to \infty} f_n d \mu = \lim_{k \to \infty} \int F_k (x) d \mu \leq \liminf_{n \to \infty} \int f_n d\mu .$$

$\qedsymbol$

We further extend the integral to complex-valued functions, just saying that

$$f:X \to \mathbb{C}$$

is integrable if its real and imaginary parts are both integrable. Then, by definition,

$$\int_E f d\mu= \int_E \Re f d\mu+ i \int_E\Im fd\mu$$

for any $E \subset X$ measurable. It follows that if $f$ is integrable then so is $\left\vert f \right\vert$. Furthermore

$$\left\vert \int_E f d \mu \right\vert \leq \int_E \left\vert f \right\vert d \mu .$$

This is obvious if $\int_E f d \mu =0$, and if not then

$$\int_E f d \mu = R e^{i\theta} R >0 , \theta \subset [0, 2 \pi) .$$

Then

$$\left\vert \int_E f d \mu \right\vert = e^{-i\theta } \int_E f d \mu$$

$$= \int_E e^{-i\theta } f d \mu$$

$$= \int_E \mathbb{R}e(e^{-i\theta }f) d \mu$$

$$\leq \int_E \left\vert \mathbb{R}e (e^{-i\theta } f) \right\vert d \mu$$

$$\leq \int_E \left\vert e^{-i\theta } f \right\vert d \mu = \int_E \left\vert f \right\vert d \mu .$$

The other important convergence result for integrals is Lebesgue's Dominated convergence theorem.

Proof. First we can make the sequence $f_n(x)$ converge by changing all the $f_n(x)$'s to zero on a set of measure zero outside which they converge. This does not change the conclusions. Moreover, it suffices to suppose that the $f_n$ are real-valued. Then consider

$$h_k = g-f_k \geq 0 .$$

Now, $\liminf_{k\to\infty}h_k=g-f$ by the convergence of $f_n;$ in particular $f$ is integrable. By monotone convergence and Fatou's lemma

$$\int (g-f)d\mu=\int \liminf_{k \to \infty} h_k d \mu \leq \liminf_{k\to\infty} \int (g-f_k) d \mu$$

$$= \int g d \mu - \limsup_{k \to \infty} \int f_k d \mu .$$

Similarly, if $H_k=g+f_k$ then

$$\int(g+f)d\mu=\int \liminf_{k \to \infty} H_k d \mu \leq \int g d \mu + \liminf_{k \to \infty} \int f_k d\mu .$$

It follows that

$$\limsup_{k\to\infty} \int f_k d \mu \leq \int f d \mu \leq \liminf_{k\to\infty} \int f_k d \mu.$$

Thus in fact

$$\int f_k d \mu \to \int f d \mu .$$

$\qedsymbol$

Having proved Lebesgue's theorem of dominated convergence, let me use it to show something important. As before, let $\mu$ be a positive measure on $X$. We have defined $L^1(X,\mu)$; let me consider the more general space $L^p (X, \mu)$. A measurable function

$$f:X \to \mathbb{C}$$

is said to be `$L^p$', for $1 \leq p< \infty$, if $\left\vert f \right\vert^p$ is integrable⁶, i.e.,

$$\int _X\left\vert f \right\vert^p d \mu < \infty .$$

As before we consider equivalence classes of such functions under the equivalence relation

$$f \sim g \Leftrightarrow \mu \left\{ x; (f-g) (x) \neq 0 \right\}=0 .$$ (32)

We denote by $L^p (X, \mu)$ the space of such equivalence classes. It is a linear space and the function

$$\Vert f \Vert _p = \left( \int_X \left\vert f \right\vert^p d \mu \right)^{1/p}$$ (33)

is a norm (we always assume $1 \leq p< \infty$, sometimes $p=1$ is excluded but later $p= \infty$ is allowed). It is straightforward to check everything except the triangle inequality. For this we start with

Proof. If $b=0$ this is easy. So assume $b>0$ and divide by $b$. Taking $t=a/b$ we must show

$$t^{\gamma} \leq \gamma t + 1- \gamma , 0 \leq t , 0< \gamma <1 .$$ (34)

The function $f(t) = t^{\gamma}- \gamma t$ is differentiable for $t>0$ with derivation $\gamma t^{\gamma -1}- \gamma$, which is positive for $t<1$. Thus $f(t) \leq f(1)$ with equality only for $t=1$. Since $f(1)=1- \gamma$, this is (3.13), proving the lemma. $\qedsymbol$

We use this to prove Hölder's inequality

Proof. If $\Vert f \Vert _p =0$ or $\Vert g \Vert _q =0$ the result is trivial, as it is if either is infinite. Thus consider

$$a= \left\vert \frac{f(x)}{\Vert f \Vert _p} \right\vert^p , b= \left\vert \frac{g(x)}{\Vert g \Vert _q} \right\vert^q$$

and apply (3.12) with $\gamma = \frac{1}{p}$. This gives

$$\frac{\left\vert f(x) g(x) \right\vert}{\Vert f \Vert _p \Vert g ... ...f \Vert _p^p} + \frac{\left\vert g(x) \right\vert^q}{q \Vert g \Vert _q^q} .$$

Integrating over $X$ we find

$$\frac{1}{\Vert f \Vert _p \Vert g \Vert _q} \int_X \left\vert f(x) g(x) \right\vert d \mu$$

$$\leq \frac{1}{p} + \frac{1}{q} =1 .$$

Since $\left\vert \int_X fg d \mu \right\vert \leq \int_X \left\vert fg \right\vert d \mu$ this implies (3.14).

$\qedsymbol$

The final inequality we need is Minkowski's inequality.

Proof. The case $p=1$ you have already done. It is also obvious if $f+g=0$ a.e.. If not we can write

$$\left\vert f+g \right\vert^p \leq \left( \left\vert f \right\vert + \left\vert g \right\vert \right) \left\vert f+g \right\vert^{p-1}$$

and apply Hölder's inequality, to the right side, expanded out,

$$\int \left\vert f+g \right\vert^p d \mu \leq \left( \Vert f \V... ... \left( \int \left\vert f+g \right\vert^{q(p-1)} d \mu \right)^{1/q} .$$

Since $q(p-1)=p$ and $1- \frac{1}{q} = 1/p$ this is just (3.15). $\qedsymbol$

So, now we know that $L^p (X, \mu)$ is a normed space for $1 \leq p< \infty$. In particular it is a metric space. One important additional property that a metric space may have is completeness, meaning that every Cauchy sequence is convergent.

Proof. We need to show that a given Cauchy sequence $\left\{ f_n \right\}$ converges in $L^p (X, \mu)$. It suffices to show that it has a convergent subsequence. By the Cauchy property, for each $k$ $\exists n=n(k)$ s.t.

$$\Vert f_n - f_{\ell} \Vert _p \leq 2^{-k} \forall \ell \geq n .$$ (35)

Consider the sequence

$$g_1=f_1 , g_k = f_{n(k)} - f_{n(k-1)} , k>1 .$$

By (3.16), $\Vert g_k \Vert _p\le2^{-k},$ for $k>1,$ so the series $\sum_{k}\Vert g_k\Vert _p$ converges, say to $B< \infty.$ Now set

$$h_n(x) = \sum^n_{k=1} \left\vert g_k (x) \right\vert , n \geq 1 , h(x)= \sum^{\infty}_{k=1} g_k(x).$$

Then by the monotone convergence theorem

$$\int_X h^p d \mu = \lim_{n \to \infty} \int_X \left\vert h_n \right\vert^p d \mu \leq B^p ,$$

where we have also used Minkowski's inequality. Thus $h \in L^p (X, \mu)$, so the series

$$f(x)= \sum^{\infty}_{k=1} g_k (x)$$

converges (absolutely) almost everywhere. Since

$$\left\vert f(x) \right\vert^p = \lim_{n \to \infty} \left\vert \sum^n_{k=1} g_k \right\vert^p \leq h^p$$

with $h^p \in L'(X, \mu)$, the dominated convergence theorem applies and shows that $f \in L^p(X, \mu)$. Furthermore,

$$\sum^{\ell}_{k=1} g_k (x) = f_{n(\ell)} (x) \hbox{ and } \left\vert f(x) - f_{n( \ell)} (x) \right\vert^p \leq (2h(x))^p$$

so again by the dominated convergence theorem,

$$\int_X \left\vert f (x) - f_{n( \ell)}(x) \right\vert^p \to 0 .$$

Thus the subsequence $f_{n( \ell)} \to f$ in $L^p (X, \mu)$, proving its completeness. $\qedsymbol$

Next I want to return to our starting point and discuss the Riesz representation theorem. There are two important results in measure theory that I have not covered -- I will get you to do most of them in the problems -- namely the Hahn decomposition theorem and the Radon-Nikodym theorem. For the moment we can do without the latter, but I will use the former.

So, consider a locally compact metric space, $X$. By a Borel measure on $X$, or a signed Borel measure, we shall mean a function on Borel sets

$$\mu :\mathcal{B}(X) \to \mathbb{R}$$

which is given as the difference of two finite positive Borel measures

$$\mu (E) = \mu_1 (E) - \mu_2 (E) .$$ (36)

Similarly we shall say that $\mu$ is Radon, or a signed Radon measure, if it can be written as such a difference, with both $\mu_1$ and $\mu_2$ finite Radon measures. See the problems below for a discussion of this point.

Let $M(X)$ denote the set of finite Radon measures on $X$. This is a normed space with

$$\Vert \mu \Vert _1 = \inf (\mu_1(X) + \mu_2(X) )$$ (37)

with the infimum over all Radon decompositions (3.17). Each signed Radon measure defines a continuous linear functional on $\mathcal{C}_0 (X)$:

$$\int \cdot \, d \mu : \mathcal{C}_0 (X) \ni f \longmapsto \int_X f \cdot \, d \mu \, .$$ (38)

Thus the dual space of $\mathcal{C}_0 (X)$ is $M(X)$ - at least this is how such a result is usually interpreted

$$(\mathcal{C}_0(X))' = M(X) \, ,$$ (39)

see the remarks following the proof.

Proof. We have done half of this already. Let me remind you of the steps.

We started with $u \in (\mathcal{C}_0(X))'$ and showed that $u=u_+-u_-$ where $u_{\pm}$ are positive continuous linear functionals; this is Lemma 1.5. Then we showed that $u \geq 0$ defines a finite positive Radon measure $\mu$. Here $\mu$ is defined by (1.11) on open sets and $\mu (E) = \mu^*(E)$ is given by (1.12) on general Borel sets. It is finite because

$$\mu(X) = \sup \left\{ u(f) ; 0 \leq f \leq 1 , \operatorname{supp} f \Subset X , f \in C(X) \right\}$$ (40)

$$\leq \Vert u \Vert .$$

From Proposition 2.8 we conclude that $\mu$ is a Radon measure. Since this argument applies to $u_{\pm}$ we get two positive finite Radon measures $\mu_{\pm}$ and hence a signed Radon measure

$$\mu = \mu_+ - \mu_- \in M(X) .$$ (41)

In the problems you are supposed to prove the Hahn decomposition theorem, in particular in Problem 14 I ask you to show that (3.22) is the Hahn decomposition of $\mu$ -- this means that there is a Borel set $E \subset X$ such that $\mu_- (E) =0 , \mu_+(X\setminus E)=0$.

What we have defined is a linear map

$$(\mathcal{C}_0(X))' \to M(X),\ u \longmapsto \mu \, .$$ (42)

We want to show that this is an isomorphism, i.e., it is $1-1$ and onto.

We first show that it is $1-1$. That is, suppose $\mu =0$. Given the uniqueness of the Hahn decomposition this implies that $\mu_+ = \mu_- =0$. So we can suppose that $u \geq 0$ and $\mu = \mu_+ =0$ and we have to show that $u=0$; this is obvious since

$$\begin{gathered}\mu (X) = \sup \left\{ u(f) ; \operator... ... \Rightarrow u(f)=0 \hbox{ for all such } f . \end{gathered}$$ (43)

If $0 \leq f \in C(X)$ and $\operatorname{supp} f \Subset X$ then $f' = f/ \Vert f \Vert _{\infty}$ is of this type so $u(f)=0$ for every $0 \leq f \in C(X)$ of compact support. From the decomposition of continuous functions into positive and negative parts it follows that $u(f)=0$ for every $f$ of compact support. Finally $\mathcal{C}_0 (X)$ is the closure of the space of continuous functions of compact support so by the assumed continuity of $u,$ $u=0.$

So it remains to show that every finite Radon measure on $X$ arises from (3.23). We do this by starting from $\mu$ and constructing $u$. Again we use the Hahn decomposition of $\mu$, as in (3.22)⁷. Thus we assume $\mu \geq 0$ and construct $u$. It is obvious what we want, namely

$$u(f) = \int_X f \, d \mu \, , \ f \in \mathcal{C}_c(X) \, .$$ (44)

Here we need to recall from Proposition 2.11 that continuous functions on $X$, a locally compact metric space, are (Borel) measurable. Furthermore, we know that there is an increasing sequence of simple functions with limit $f$, so

$$\left\vert \int_X f d \mu \right\vert \leq \mu (X) \cdot \Vert f \Vert _{\infty} .$$ (45)

This shows that $u$ in (3.25) is continuous and that its norm $\Vert u \Vert \leq \mu (X)$. In fact

$$\Vert u \Vert = \mu (X) .$$ (46)

Indeed, the inner regularity of $\mu$ implies that there is a compact set $K \Subset X$ with $\mu (K) \geq \mu (X) - \frac{1}{n}$; then there is $f \in \mathcal{C}_c (X)$ with $0 \leq f \leq 1$ and $f=1$ on $K$. It follows that $\mu (f) \geq \mu (K) \geq \mu(X) - \frac{1}{n}$, for any $n$. This proves (3.27).

We still have to show that if $u$ is defined by (3.25), with $\mu$ a finite positive Radon measure, then the measure $\tilde{\mu}$ defined from $u$ via (3.25) is precisely $\mu$ itself.

This is easy provided we keep things clear. Starting from $\mu \geq 0$ a finite Radon measure, define $u$ by (3.25) and, for $U\subset X$ open

$$\tilde{\mu} (U) = \sup \left\{ \int_X f d \mu , 0 \leq f \leq 1 , f \in C(X) , \operatorname{supp} (f) \Subset K \right\} .$$ (47)

By the properties of the integral, $\tilde{\mu} (U) \leq \mu (U)$. Conversely if $K \Subset U$ there exists an element $f \in \mathcal{C}_c (X)$, $0 \leq f \leq 1$, $f=1$ on $K$ and $\operatorname{supp} (f) \subset U$. Then we know that

$$\tilde{\mu} (U) \geq \int_X f d \mu \geq \mu (K) .$$ (48)

By the inner regularity of $\mu$, we can choose $K \Subset U$ such that $\mu (K) \geq \mu (U) - \epsilon$, given $\epsilon >0$. Thus $\tilde{\mu} (U) = \mu (U)$.

This proves the Riesz representation theorem, modulo the decomposition of the measure - which I will do in class if the demand is there! In my view this is quite enough measure theory. $\qedsymbol$

Notice that we have in fact proved something stronger than the statement of the theorem. Namely we have shown that under the correspondence $u \longleftrightarrow \mu$,

$$\Vert u \Vert = \left\vert \mu \right\vert (X) = : \Vert \mu \Vert _1 .$$ (49)

Thus the map is an isometry.