Convergence of Measurable Functions

Recall that a function $f:X\to \mathbb{R}$ is $\mathcal{S}$-measurable if the preimage of every Borel set $B$ is in the $\sigma$-algebra $\mathcal{F}$:

$$\forall B\in \mathcal{B}{f}^{-1}(B)\in \mathcal{S}$$

We review the usual pointwise and uniform convergence.

Def Suppose $X$ is a set, $\{f_n{\}}_{n\ge 1}$ is a sequence of functions from $X$ to $\mathbb{R}$, and $f$ is a function from $X$ to $\mathbb{R}$.

• The sequence $\{f_n{\}}_{n\ge 1}$ converges pointwise on $X$ to $f$ if

$$\underset{n\to \infty }{\lim }{f}_n(x)=f(x)$$

for each $x\in X$. That is, $\forall x\in X$ and $\forall ϵ>0$ there exists $N_x$ such that

$$|f_n(x)-f(x)|<ϵ\forall n\ge N_x$$

• The sequence $\{f_n{\}}_{n\ge 1}$ converges uniformly on $X$ to $f$ if for every $ϵ>0$ there exists an $N$ such that

$$|f_n(x)-f(x)|<ϵ\forall n\ge N,\forall x\in X$$

Like the difference between continuity and uniform continuity, the difference lies in the order of the quantifiers. Clearly uniform convergence implies pointwise convergence, a basic counterexample is the sequence

$$f_n(x)=1_{[-\frac{1}{n},+\frac{1}{n}]}(x)$$

which converges pointwise the the function $f(0)=1$ and $f(x)=0$ if $x\ne 0$, but not uniformly.

Also if $f_n\to f$ and $f_n\in C$ doesn not imply $f\in C$, a counterexample is just a smooth version of the previous characteristic function. On the other hand, if $f_n\rightrightarrows f$ then also $f\in C$.

Proposition If a sequence $\{f_n{\}}_{n\ge 1}$of continuous function converges uniformly to a function $f$, then $f\in C(X)$. Dim Let $x\in X$, for all $ϵ>0$

$$|f(x)-f(y)|\le |f_n(x)-f(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$

since we have uniform convergence there exists $N$ such that $\forall n\ge N$ the first and last term on the right are less than $\frac{ϵ}{3}$. Since $f_n$ is continuous, there exists a $\delta >0$ such that if $|x-y|<\delta$ the second term is less than $\frac{ϵ}{3}$. $\square$.

In a finite measure space, pointwise convergence is almost uniform convergence, in the sense that it converges uniformly except on a set of arbitrary small measure.

Theorem (Egorov) Let $(X,\mathcal{S},\mu )$ be a finite measure space. Suppose $\{f_n{\}}_{n\ge 1}$ is a sequence of $\mathcal{S}$-measurable functions from $X$ to $\mathbb{R}$ such that $f_n\to f$. Then for every $ϵ>0$ there exists a set $E\subseteq X$ such that $\mu (X\setminus E)<ϵ$ and $f_n\rightrightarrows f$ on $E$. Proof The definition of pointwise convergence can rewritten as: for every $k\ge 1$

$$\bigcup _{N=1}^{\infty }\bigcap _{n=N}^{\infty }\left\{x\in X:|f_n(x)-f(x)|<\frac{1}{k}\right\}=X$$ $$\bigcup _{N=1}^{\infty }{A}_{N,k}=X$$

where we have defined the sets $A_{N,k}$, which are in $\mathcal{S}$ since $f_n-f$ is still $\mathcal{S}$-measurable.

Since this is an increasing sequence of sets, the Continuity of mesure implies

$$\underset{N\to \infty }{\lim }\mu (A_{N,k})=\mu (X)$$

thus there exists $N_k$ such that

$$\mu (X)-\mu (A_{N,k})<\frac{ϵ}{2^k}\forall N\ge N_k$$

Sia

$$E=\bigcap _{k=1}^{\infty }{A}_{N_k,k}$$

then

$$\mu (X\setminus E)=\mu \left(X\setminus \bigcap _{k=1}^{\infty }{A}_{N_k,k}\right)=\mu \left(\bigcup _{k=1}^{\infty }X\setminus A_{N_k,k}\right)$$ $$\le \sum _{k=1}^{\infty }\mu (X\setminus A_{N_k,k})<ϵ$$

We must verify uniform convergence in $E$, for every $ϵ>0$ we can finde $k\ge 1$ such that $\frac{1}{k}\le ϵ$, by definition $E\subseteq A_{N_k,k}$ so

$$|f_n(x)-f(x)|<\frac{1}{k}\le ϵ\forall n\ge N_k$$

for all $x\in E$, hence $f_n\rightrightarrows f$. $\square$

Every measurable function can be approximated from below by simple functions, in the sense that we can find an increasing sequence of simple function such that ${\varphi }_n\to f$, moreover if $f$ is bounded ${\varphi }_n\rightrightarrows f$.

Proposition Let $(X,\mathcal{S})$ be a measurable space and $f:X\to [-\infty ,+\infty ]$ be $\mathcal{S}$-measurable. Then there exists a sequence ${{\varphi }_n}_{n\ge 1}$ of simple functions such that

• $|{\varphi }_n(x)|\le |{\varphi }_{n+1}(x)|\le f(x)|$ for all $n$ and all $x\in X$.
• ${\varphi }_n\to f$ (pointwise)
• if $f$ is bounded, ${\varphi }_n\rightrightarrows f$. Proof We construct the usual stair functions:

$${\varphi }_n(x)=\{\begin{array}{l}n\text{ if }f(x)>k\\ -n\text{ if }f(x)<-k\\ \frac{\lfloor 2^{n}f(x)\rfloor }{2^n}\text{ otherwise }\end{array}$$

it’s a simple function, and the definition implies that where $f(x)\in [-n,n]$

$$|{\varphi }_n(x)-f(x)|\le \frac{1}{2^n}$$

which shows the second and third claim. $\square$

The next result is surprising, a Borel function is almost continuous, in the sense that there exists a large closed set on where continuous.

Theorem (Luzin) Suppose $f:\mathbb{R}\to \mathbb{R}$ is a Borel measurable function. Then for every $ϵ>0$ there exists a closed set $F\subseteq \mathbb{R}$ such that $\lambda (\mathbb{R}\setminus F)<ϵ$ and $g{|}_F$ is continuous on $F$. Proof The claim is clear if $g$ is a simple functions: any Borel set can be approximated from the inside with a closed set, and from the outside with an open set. If we restrict the function on the disjoint close sets, $g$ is continuous. Now by the previous proposition we can find a sequence of simple function that convergences pointwise to $g$, restrict the domain such that they are continuous, then use Egorov’s theorem and the fact that uniform convergence preserves continuity.