Chapter 4 Mesurable funtions

A big part of measure theory is the study of functions which are compatible with the measure spaces. We begin with a basic definition which will be satisfied by all the functions we are interested in.

Definition 4.1 (Mesasurable functions) If $(E, \mathcal{E})$ and $(F, \mathcal{F})$ are two measurable spaces and $f$ is a function $E \rightarrow F$, then we say $f$ is measurable if for every $A \in \mathcal{F}$ we have $f^{-1}(A) \in \mathcal{E}$.

Lemma 4.1 Suppose that $\mathcal{A} \subset \mathcal{F}$ is such that $\sigma(\mathcal{A})= \mathcal{F}$. If $f$ is a function such that for every $A \in \mathcal{A}$ we have $f^{-1}(A) \in \mathcal{E}$ then $f$ is measurable.

Proof. First we note that

\[ f^{-1}\left( \bigcup_i A_i \right) = \bigcup_i f^{-1}(A_i), \]

and

\[ f^{-1}(B \setminus A) = f^{-1}(B) \setminus f^{-1}(A). \]

Now if we consider $\{ A \in \mathcal{F} \, :\, f^{-1}(A) \in \mathcal{E}\}$ then this is a $\sigma$-algebra, as $\mathcal{E}$ is a $\sigma$-algebra and $f^{-1}$ preserves set operations. Therefore, $\{ A \in \mathcal{F} \, :\, f^{-1}(A) \in \mathcal{E}\}$ is a $\sigma$-algebra containing $\mathcal{A}$ therefore $\mathcal{F} \subseteq \{ A \in \mathcal{F} \, :\, f^{-1}(A) \in \mathcal{E}\}$ so $f$ is measurable.

Remark. In particular note that the above lemma means that whenever we have $f: E \rightarrow \mathbb{R}$ and $\mathbb{R}$ is equipped with the Borel $\sigma$ algebra, we know that $f$ is measurable if $f^{-1}((-\infty, b])$ is a measurable set for every $b$.

Lemma 4.2 If $E, F$ are topological spaces, equipped with their Borel $\sigma$-algebras, and we have $f:E \rightarrow F$ is a continuous map then $f$ is measurable.

Proof. This is on the exercise sheet and a solution on the solution sheet.

Lemma 4.3 If $(E, \mathcal{E}), (F, \mathcal{F})$ and $(G, \mathcal{G})$ are all measurable spaces and $f : E \rightarrow F$ and $g: F \rightarrow G$ are measurable then so is $g \circ f$.

Proof. Take any set $A \in \mathcal{G}$ then $(g\circ f)^{-1}(A) = \{ x \in E \,:\, g(f(x)) \in A\}$. Let us call $B = \{y \in F \,:\, g(y) \in A\} = g^{-1}(A)$ then $(g \circ f)^{-1}(A) = \{ x \in E \,:\, f(x) \in B\} = f^{-1}(B)$. Then as $g$ is measurable and $A \in \mathcal{G}$ then $B \in \mathcal{F}$. In the same way as $f$ is measurable and $B \in \mathcal{F}$ then $f^{-1}(B) \in \mathcal{E}$. As $f^{-1}(B) = (g \circ f)^{-1}(A)$ this shows that $(g \circ f)^{-1}(A) \in \mathcal{E}$ for every $A \in \mathcal{G}$ and hence $g \circ f$ is measurable.

Lemma 4.4 Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a monotone function. Then $f$ is measurable with respect to the Borel $\sigma$-algebra.

Proof. Suppose without loss of generality that $f$ is increasing then $f^{-1}((-\infty, b])$ is $\emptyset, (-\infty, \infty)$ or $(-\infty, a)$ or $(-\infty, a]$ for some $a$. All these possibilities are Borel measurable sets.

Lemma 4.5 If $f_n$ is a sequence of measurable function taking values in $(\mathbb{R}, \mathcal{B}(\mathbb{R})$ then the following functions are all measurable:

$-f_1$
$\lambda f_1$ for $\lambda >0$ a fixed contant.
$f_1 \wedge f_2$
$f_1 \vee f_2$
$f_1+f_2$,
$f_1 f_2$,
$\sup_n f_n$,
$\inf_n f_n$,
$\limsup_n f_n$,
$\liminf_n f_n$.

Proof. We only show two result. The rest are on the assignment.

In order to show that any of these functions are measureable we want to look at $f^{-1}((-\infty, b])$ or a similar set. $(f_1 \vee f_2)^{-1}((-\infty, b]) = \{ x \,:\, \max\{f_1(x), f_2(x)\} \leq b\} = \{ x \,:\, f_1(x) \leq b \, \mbox{and} \, f_2(x) \leq b\} = \{ x \,:\, f_1(x) \leq b \} \cap \{x \,:\, f_2(x) \leq b\} = f_1^{-1}((-\infty, b]) \cap f_2^{-1}((-\infty, b])$. Now since $f_1$ and $f_2$ are both measureable the sets $f_1^{-1}((-\infty, b])$ and $f_2^{-1}((-\infty, b])$ are both measurable. We also know that the intersection of two measurable sets is measurable.

$(f_1+f_2)^{-1}((b,\infty)) = \{ x \,:\, f_1(x) + f_2(x) > b \}$. Now if $f_1(x) > b-f_2(x)$ then there exists a $q \in \mathbb{Q}$ such that $f_1(x)> q > b - f_2(x)$. Let us define the set $A= \bigcup_{q \in \mathbb{Q}} \{ x \,:\, f_1(x) > q\} \cap \{ x\,:\, f_2(x) > b-q\}$. Since $f_1,f_2$ are both measurable $A$ is a countable union of measurable sets so measurable. We can also see that if $x \in A$ then $f_1(x) + f_2(x) > b$ and our observation shows that in fact $A= \{ x \,:\, f_1(x) + f_2(x) > b \}$. Therefore, $f_1+f_2$ is measurable.

Definition 4.2 (Image measure) We can use a measurable function $f$ to define an image measure. Suppose $\mu$ is a measure on $(E, \mathcal{E})$ and $f$ is a measurable function $(E, \mathcal{E}) \rightarrow (F, \mathcal{F})$ then we can define a new measure $\nu$ by saying that

\[ \nu(A) = \mu(f^{-1}(A)), \]

for every $A \in \mathcal{F}$. We write $\nu = \mu \circ f^{-1}$.

We can use the notion of image measure to construct further measures from Lebesgue measure.

Lemma 4.6 Suppose $g: \mathbb{R} \rightarrow \mathbb{R}$ and that $g$ is non-constant, right-continuous and non-decreasing. Let us define $g(-\infty) = \lim_{x \rightarrow -\infty} g(x)$ and $g(\infty) = \lim_{x \rightarrow \infty} g(x)$ and let us call the interval $I:= (g(-\infty),g(\infty))$ (this might be the whole of $\mathbb{R}$. Define a partial inverse to $g$ by $f: I \rightarrow \mathbb{R}$ by

\[ f(x) = \inf \{y \,:\, x \leq g(y)\}. \]

Then $f$ is left-continuous and non-decreasing and $f(x) \leq y$ if and only if $x \leq g(y)$.

Proof. Fix $x \in I$ and consider the set $J_x = \{ y \in \mathbb{R}\,:\, x \leq g(y)\}$ by definition of $I$ we know that $J_x$ is non empty and is not the whole of $\mathbb{R}$ (this shows that $f$ is well defined). As $g$ is non-decreasing, if $y \in J_x$ and $y' \geq y$, then $y' \in J_x$. As $g$ is right-continuous, if $y_n \in J_x$ and $y_n \downarrow y$ then $y \in J_x$ (noting the $\leq$ sign in $J_x$). Now using this we have that if $x \leq x'$ then $J_x \supseteq J_{x'}$ so $f(x) \leq f(x')$. We also have that if $x_n \uparrow x$ then $J_x = \bigcap_n J_{x_n}$, so $f(x_n) \rightarrow f(x)$.

Theorem 4.1 Let $g$ be a non-constant, right-continuous and non-decreasing function from $\mathbb{R} \rightarrow \mathbb{R}$. There exists a unique Radon measure on $\mathbb{R}$ such that for all $a,b \in \mathbb{R}$ with $a < b$

\[ \mathrm{d}g((a,b]) = g(b) - g(a). \]

We call this measure the Lebesgue Steitjles measure associated with $g$. Furthermore, every Radon measure on $\mathbb{R}$ can be represented this way.

Proof. Define $I$ and $f$ as in the Lemma above. Then we can construct $\mathrm{d}g$ as the image measure of Lebesgue measure on $I$. That is to say we can let $\mathrm{d}g = \lambda \circ f^{-1}$. If this is the case then

\[ \mathrm{d}g ((a,b]) = \lambda \left(\{ x \, :\, f(x) > a, f(x) \leq b \} \right) = \lambda ((g(a), g(b)]) = g(b) - g(a). \]

The standard argument for uniqueness of measures (as for that of Lebesgue measure) gives uniqueness of this measure.

Finally, if $\nu$ is a Radon measure on $\mathbb{R}$ then we can define a function $g$, by

\[ g(y) = \nu((0,y]), \quad y \geq 0, \quad g(y) = -\nu((y,0]), \quad y<0. \]

Then $\nu = \mathrm{d}g$ by uniqueness.

4.0.1 Random variables and the measure theoretic formulation of probability - NONEXAMINABLE

The structure of measure theory allows us to put probability theory on a firm footing. This section is mainly as an example and for those of you who are interested in probability.

Definition 4.3 We call a measure space $(\Omega, \mathcal{F}, \mathbb{P})$ a probability space (and use different letters for the different bits) if $\mathbb{P}(\Omega) = 1$. In this setting we still have $\Omega$ is a set, $\mathcal{F}$ is a $\sigma$-algebra and $\mathbb{P}$ is a measure. We call $\mathbb{P}$ a probability measure.

The set $\Omega$ is not really described and we view it as the space of all possible outcomes. We call $A \subset \Omega$ and event if $A \in \mathcal{A}$, and $\mathbb{P}(A)$ the probability of an event happening.

Definition 4.4 A random variable, $X$ is a measurable function from a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ to another measurable space $(E, \mathcal{A})$.

Under this way of writng things, with $B \in \mathcal{A}$, we have $\mathbb{P}(X \in B) = \mathbb{P}(\{ \omega \in \Omega \,:\, X(\omega) \in B\}) = \mathbb{P}( X^{-1}(B))$, where $X^{-1}(B) \in \mathcal{F}$ as $X$ is measurable. We call $X^{-1}(B)$ the event that $X \in B$. When working with probability people usually suppress the argument $\omega$.

Definition 4.5 The law of a random variable, $X$ is the image measure of $\mathbb{P}$ under the measurable function $X$. I.e. if $X: (\Omega, \mathcal{F}, \mathbb{P}) \rightarrow (E, \mathcal{A})$ then we define a measure $\mu_X$ on $(E, \mathcal{A})$ by $\mu_X(B) = \mathbb{P}(X \in B)$.

Remark. The law of a random variable is an object which allows us to understand both probability densities and discrete probability distributions in the same way.

If $X$ takes values in $\mathbb{R}$ then the distribition function of $X$, $F_X(x) = \mu_X((-\infty, x])$. If $X$ has density $f(x)$ then $\mu_X$ is equal to the measure given by $\mu_X(A) = \int_A f(x) \mathrm{d}x$, though we still don’t know how to integrate over complicated sets.

4.1 Convergence of measurable functions

Definition 4.6 (Almost everywhere / Almost surely) We use the short hand almost everywhere (or almost surely in a probability space) to discuss properties that are true everywhere except a measure zero set.

Definition 4.7 (Convergence almost everywhere) Let $(E, \mathcal{E}, \mu)$ be a measureable space. A sequence of measureable functions, $(f_n)_{n \geq 1}: E \rightarrow F$, converges almost everywhere to $f$ if

\[ \mu \left( \{ x \in E \,:\, f_n(x) \not\to f(x) \} \right) = 0 \]

Definition 4.8 (Convergence in measure) Let $(E, \mathcal{E}, \mu)$ be a measureable space. A sequence of real valued measureable functions, $(f_n)_{n \geq 1}: E \rightarrow F$, converges in measure to $f$ if for every $\epsilon > 0$

\[ \mu \left( \{ x \, :\, |f(x) - f_n(x)| > \epsilon \} \right) \rightarrow 0, \quad \mbox{as}\, n \rightarrow \infty. \]

Example 4.1 The sequence of functions $f_n(x) = x^n$ converges to $0$ Lebesgue almost everywhere on $[0,1]$, and in measure, but it doesn’t converge pointwise as it doesn’t converge at $x=1$.

Example 4.2 The- sequence of functions $f_n(x) = 1_{[n,n+1]}(x)$ converges to 0 Lebesgue almost everywhere (in fact everywhere) but not in measure.

Example 4.3 Consider the sequence of functions $f_1 = 1_{[0,1/2)}, f_2 = 1_{[1/2, 1)}, f_3 = 1_{[0,1/4)}, f_4 = 1_{[1/4, 1/2)}, f_5= 1_{[1/2, 3/4)}, f_6 = 1_{[3/4,1)}, f_7 = 1_{[0,1/8)}, f_8 = 1_{[1/8, 1/4)} \dots$ then $f_n$ converges to 0 in measure, but $f_n(x)$ does not converge for any $x$.

We can prove a quasi-equivalence between these two notions of measure. Before we do this we need to introduce a very useful lemma, the Borel-Cantelli Lemma. We introduce it here as it is used to prove the following theorem but it is a useful tool to have whilst doing measure theory, particularly probability theory. First let us also introduce some more notation

Definition 4.9 Let $(A_n)_n$ be a sequence of measurable sets then we have the following names

\[ \limsup_n A_n = \bigcap_n \bigcup_{m \geq n} A_m = \{ A_n \, \mbox{infinitely often}\,\}, \]

and

\[ \liminf_n A_n = \bigcup_n \bigcap_{m \geq n} A_m = \{ A_n \, \mbox{eventually}\,\}. \]

The last names are more common when the $A_n$ are events in a probability space.

Lemma 4.7 (First Borel-Cantelli Lemma) Let $(E, \mathcal{E}, \mu)$ be a measure space. Then if $\sum_n \mu(A_n) < \infty$ it follows that $\mu(\limsup_n A_n) = 0)$.

Proof. For any $n$ we have

\[ \mu(\limsup_n A_n) \leq \mu \left( \bigcup_{m \geq n} A_m\right) \leq \sum_{m \geq n} \mu(A_m). \]

Then the right hand side goes to zero as $n \rightarrow \infty$, so $\mu(\limsup_n A_n) = 0$.

Lemma 4.8 (Second Borel-Cantelli Lemma - NONEXAMINABLE) Let $(E, \mathcal{E}, \mu)$ be a probability space ($\mu(E) =1$). Then suppose that the events are all inedepenent and that $\sum_n \mu(A_n) = \infty$ then it will follow that $\mu(\limsup_n A_n) =1$.

Proof. First we note that $\mu(A_i^c \cap A_j^c) = \mu ((A_i \cup A_j)^c) = 1 - \mu(A_i \cup A_j) = 1 - \mu(A_i) - \mu(A_j)+ \mu(A_i)\mu(A_j) = (1-\mu(A_i))(1-\mu(A_j))$. We use the inequality $1-a \leq e^{-a}$. Let $a_n = \mu(A_n)$ then

\[ \mu \left( \bigcap_{m=n}^N A_m^c \right) = \Pi_{m=n}^N (1-a_m) \leq \exp \left( - \sum_{m=n}^N a_m \right) \rightarrow 0, \quad \mbox{as}\, N \rightarrow 0. \]

Therefore, $\mu \left( \bigcap_{m \geq n} A_m^c \right) = 0$ for every $n$. So $\mu(\limsup_n A_n ) = 1- \mu(\bigcup_n \bigcap_{m \geq n} A_m^c) = 1$.

Theorem 4.2 Let $(E, \mathcal{E}, \mu)$ be a measure space and $(f_n)_n$ be a sequence of measurable functions. Then we have the following:

Suppose that $\mu(E) < \infty$ and that $f_n \rightarrow 0$ almost everywhere, then $f_n \rightarrow 0$ in measure.
If $f_n \rightarrow 0$ in measure then there exists some subsequence $(n_k)_k$ such that $f_{n_k} \rightarrow 0$ almost everywhere.

Proof. Suppose that $f_n \rightarrow 0$ almost everywhere. Then

\[ \mu(\{ x\,:\, |f_n(x)| \leq \epsilon \}) \geq \mu \left( \bigcap_{m \geq n} \{ x \,:\, |f_m(x)| \leq \epsilon\}\right) \uparrow \mu \left(|f_n| \leq \epsilon \, \mbox{eventually} \right) \geq \mu(\{ x \,:\, f_n(x) \rightarrow 0\}) = \mu(E), \]

therefore,

\[ \mu(\{ x\,:\, |f_n(x)|> \epsilon) = \mu(E) - \mu(\{ x\,:\, |f_n(x)| \leq \epsilon \}) \rightarrow 0. \]

Now suppose that $f_n \rightarrow 0$ in measure. We can find a subsequence $n_k$ such that

\[ \mu(\{ x \,:\, |f_{n_k}(x)| > 1/k \}) \leq 2^{-k}. \]

Therefore

\[ \sum_k \mu(\{ x \,:\, |f_{n_k}(x)| > 1/k \})<\infty.\]

Therefore by the first Borel-Canteli lemma we have that

\[ \mu \left(\{ x\,:\, |f_{n_k}(x)|>1/k \, \mbox{infinitely often}\} \right) = 0. \]

Therefore $f_{n_k} \rightarrow 0$ almost everywhere.

4.2 Egoroff’s Theorem and Lusin’s Theorem

Theorem 4.3 (Egoroff's Theorem) Let $(E, \mathcal{E}, \mu)$ be a finite measure space and $(f_n)_n$ be a sequence of real valued measurable functions on $E$. If $\mu$ is finite and if $f_n$ converges $\mu$-almost everywhere to $f$ then for each positive $\epsilon$ there is a set $A$ with $\mu(A^c)< \epsilon$, such that $f_n$ converges uniformly on $A$ to $f$.

Proof. For each $n$ let $g_n(x) = \sup_{j \geq n}|f_j(x)-f(x)|$. Then $g_n$ is a positive function which is finite almost everywhere. The sequence $(g_n)_n$ converges to 0 almost everywhere and so in measure. Therefore, for each positive integer $k$ we can find $n_k$ such that

\[ \mu \left( \{ x \,:\, g_{n_k} > 1/k \} \right) < \epsilon 2^{-k}. \]

Define sets $A_k = \{ x\,:\, g_{n_k}(x) \leq 1/k\}$ and let $A= \bigcap_k A_k$. The set $A$ has

\[ \mu(A^c) = \mu \left( \bigcup_k A_k^c \right) \leq \sum_k \mu(A_k^c) \leq \sum_k \epsilon 2^{-k} = \epsilon. \]

We want to show that $f_n$ converges uniformly to $f$ on $A$. For each $\delta$ there exists a $k$ such that $1/k < \delta$, then as $A \subseteq A_k$, we have that for every $n \geq n_k$,

\[ |f_n - f| \leq g_{n_k} \leq 1/k < \delta, \]

uniformly on all $x \in A_k$ and hence in $A$.

This proof motivates the following definition of convergence of functions, which we won’t use a lot.

Definition 4.10 (Almost uniform convergence) We say a sequence of functions $(f_n)_{n \geq 1}$ converges almost uniformly on a measure space $(E, \mathcal{E}, \mu)$ if for every $\epsilon >0$ there exists a set $A$ with $\mu(A^c)< \epsilon$ with $f_n \rightarrow f$ uniformly on $A$.

We can use Egoroff’s theorem to prove a result called Lusin’s theorem. First let us recall the definition of regularity

Definition 4.11 Let $E$ be a topological space and $\mu$ be a measure on $(E, \mathcal{B}(E))$ then say $\mu$ is regular if for every $A \in \mathcal{B}(E)$ we have

\[\mu(A) = \inf \{ \mu(U) \,:\, A \subseteq U, \mbox{$U$ is open}\},\]
\[\mu(A) = \sup \{ \mu(K) \,:\, K \subseteq A, \mbox{$K$ is compact}\}.\]

Theorem 4.4 (Lusin's Theorem) Suppose that $f$ is a measurable function and $A \subseteq \mathbb{R}^d$ is a Borel set and $\lambda(A) < \infty$ then for any $\epsilon >0$ there is a compact subset $K$ of $A$ with $\lambda(A \setminus K) < \epsilon$ such that the restriction of $f$ to $K$ is continuous.

Remark. This theorem can be generalised to locally compact Hausdorff spaces, see Cohn’s book.

Proof. Suppose first that $f$ only takes countably many values, $a_1, a_2, a_3, \dots$ on $A$ the let $A_k = \{ x \in A \,:\, f(x) = a_k\}$, by measurablility of $f$ we can see that $A_k = f^{-1}(\{a_k\})$ is measurable. We know that $A = \bigcup_n A_n$ so by continuity of measure $\lambda(\bigcup_{k=1}^n A_k) \uparrow \lambda(A)$. Since $\lambda(A) < \infty$ we have that for any $\epsilon >0$ there exists $n$ such that $\lambda(A \setminus \bigcup_{k=1}^n A_k) < \epsilon/2$. By the regularity of Lebesgue measure we can find compact subsets $K_1, \dots, K_n$ such that $\lambda(A_k \setminus K_k) \leq \epsilon/2n$ for $k=1,\dots,n$. Then let $K = \bigcup_{k=1}^n K_k$. This is a compact subset of $A$ and

\[ \lambda(A \setminus K) \leq \lambda(A\setminus \bigcup_{k=1}^n A_k) + \lambda(\bigcup_{k=1}^n A_k \setminus \bigcup_{k=1}^n K_k ) < \epsilon/2 + \epsilon/2. \]

Now $f$ restricted to $K$ is continuous since the $K_i$ are disjoint and $f$ is constant on each $K_i$.

Now we have proved the special case where $f$ takes countably many values we can use this to prove the theorem for general $f$. Let $f_n = 2^{-n} \lfloor 2^n f \rfloor$ then $2^{-n} \geq f(x)-f_n(x) \geq 0$ so $f_n(x) \rightarrow f(x)$, uniformly. Now, $f_n$ can only take countably many values, so by our special case of Lusin’s theorem there exists a $K_n \subseteq A$, compact, such that $\lambda(A \setminus K_n) \leq \epsilon 2^{-n}$, and $f_n$ is continuous on $K_n$. Now let $K_\infty = \bigcap_n K_n$, then $K_\infty$ is compact and $\lambda(A \setminus K_\infty) = \lambda (\bigcup_n(A \setminus K_n)) \leq \sum_n \epsilon 2^{-n} = \epsilon$. Now we have that $f_n$ converges uniformly to $f$ on $K_\infty$ and $f_n$ is continuous on $K_\infty$ for each $n$. As the uniform limit of continuous functions is continuous this shows that $f$ is continuous on $K_\infty$.