Measure Theory/Approximations of Measurable Functions

Let ${\displaystyle \psi =\sum _{i=1}^{n}c_{i}\mathbf {1} _{E_{i}}}$ be a simple function on the bounded interval [a,b]. Also let ${\displaystyle \varepsilon \in {\mathbb {R}}^{+}}$.

Prove that there exists a step function s and measurable set F such that

s and ${\displaystyle \psi }$ are equal on F, and F is nearly all of [a,b].

More formally, for all ${\displaystyle x\in F}$, we have ${\displaystyle \psi (x)=s(x)}$.

And ${\displaystyle \lambda ([a,b]\smallsetminus F)<\varepsilon }$.

Hint: Approximate each of ${\displaystyle E_{1},\dots ,E_{n}}$.

Let s be any step function on a closed, bounded interval, ${\displaystyle s:[a,b]\to {\mathbb {R}}}$ and ${\displaystyle s=\sum _{i=1}^{n}c_{i}\mathbf {1} _{I_{i}}}$. Also let ${\displaystyle \varepsilon \in {\mathbb {R}}^{+}}$.

Prove that there is a continuous function ${\displaystyle f:[a,b]\to {\mathbb {R}}}$ and measurable set ${\displaystyle E\subseteq [a,b]}$, such that

${\displaystyle {\text{ for all }}x\in E:f(x)=s(x),\quad {\text{ and }}\lambda ([a,b]\smallsetminus E)<\varepsilon }$

Hint: There is a finite number of discontinuities of s.

Put a small enough neighborhood around each discontinuity. Outside of these neighborhoods, make f and s equal.

Inside of these neighborhoods, interpolate a linear function from one end to the other.

Here we study two kinds of approximation results. One defines a notion in which two functions are "basically the same" (at least for the purposes of integration). This is the idea of functions being equal "almost everywhere". For functions equal almost everywhere, one may replace one function by the other and the value of the integral is unchanged.

The other kind of approximation result is that every measurable function is (1) approximately continuous, and (2) approximately step. This means that measurable functions can be replaced by continuous or step functions, and although this changes the value of the integral, it does so "not too much".

Show that the constant function 0, and the Dirichlet function ${\displaystyle \mathbf {1} _{\mathbb {Q}}}$ are equal almost everywhere.

Suppose ${\displaystyle f,g:E\to {\mathbb {R}}}$ are two functions, and f is measurable, and suppose ${\displaystyle f=g{\text{ a.e. }}}$ We will prove that therefore g is measurable.

Let ${\displaystyle a\in {\mathbb {R}}}$ and consider the sets

${\displaystyle F=f^{-1}((a,\infty ))=\{x\in E|g(x)>a\}}$

${\displaystyle G=g^{-1}((a,\infty ))=\{x\in E|f(x)>a\}}$

${\displaystyle H=\{x\in E|f(x)\neq g(x)\}}$

Justify why the following sets are measurable, in this order:

1. F

2. H

3. ${\displaystyle F\smallsetminus H}$

4. ${\displaystyle G\smallsetminus H}$

5. G

In this section we will show that, if ${\displaystyle \psi :[a,b]\to {\mathbb {R}}}$ is a simple function then there is a step function ${\displaystyle s:[a,b]\to {\mathbb {R}}}$ and a set ${\displaystyle E\subseteq [a,b]}$ such that ${\displaystyle \psi =s}$ on E and ${\displaystyle \lambda ([a,b]\smallsetminus E)<\varepsilon }$.

So let ${\displaystyle \psi =\sum _{i=1}^{n}c_{i}\mathbf {1} _{E_{i}}}$ be a simple function and let ${\displaystyle \varepsilon \in {\mathbb {R}}^{+}}$.

As we proved in a previous lesson, each ${\displaystyle E_{i}}$ is approximately an open set. That is to say, there is an open set ${\displaystyle E_{i}\subseteq U_{i}}$ such that ${\displaystyle \lambda (U_{i}\smallsetminus E_{i})<\varepsilon /n}$.

We will now show that for any measurable function ${\displaystyle f:[a,b]\to {\mathbb {R}}}$ and ${\displaystyle \varepsilon \in {\mathbb {R}}^{+}}$, there exists a continuous function ${\displaystyle g:[a,b]\to {\mathbb {R}}}$ such that

${\displaystyle |f-g|<\varepsilon }$ on a set ${\displaystyle E\subseteq [a,b]}$

and ${\displaystyle \lambda ([a,b]\smallsetminus E)<\varepsilon }$. When ${\displaystyle \varepsilon }$ is "very small" then E is almost the whole interval [a,b], and g is very close to f.

We will also prove that there exists a step function ${\displaystyle s:[a,b]\to {\mathbb {R}}}$ such that

${\displaystyle |f-s|<\varepsilon }$ on a set ${\displaystyle E\subseteq [a,b]}$

and ${\displaystyle \lambda ([a,b]\smallsetminus E)<\varepsilon }$.

First we prove that there is an M such that ${\displaystyle |f|\leq M}$ except on a set of measure less than ${\displaystyle \varepsilon /3}$.

To do so, consider the sequence of sets

${\displaystyle E_{n}=\{x\in [a,b]:|f(x)|\leq n\}}$

First show that ${\displaystyle E_{n}}$ is measurable for each ${\displaystyle 1\leq n}$.

Next show that ${\displaystyle [a,b]=\bigcup _{n=1}^{\infty }E_{n}}$.

Then use the continuity of measure to show that there is some M for which ${\displaystyle \lambda \left(\bigcup _{n=1}^{M}E_{n}\right)>b-a-\varepsilon }$. Define ${\displaystyle F=\bigcup _{n=1}^{M}E_{n}}$.

Infer that this is the desired M.

Next we will show that there is a simple function ${\displaystyle \varphi }$ such that ${\displaystyle |f-\varphi |<\varepsilon }$ except on F.

To do so, set ${\displaystyle n\in {\mathbb {N}}}$ such that ${\displaystyle {\frac {M}{n}}<\varepsilon }$ and define the sets

${\displaystyle {\begin{aligned}E_{n,i}&=\left\{x\in [a,b]:{\frac {M(i-1)}{n}}\leq |f(x)|<{\frac {Mi}{n}}\right\}\\&=(|f|)^{-1}([M(i-1)/n,Mi/n)),\quad {\text{ for }}1\leq i\leq n\end{aligned}}}$

and then define the function

${\displaystyle \varphi =\sum _{i=1}^{n}\left(\inf _{E_{n,i}}f\right)\mathbf {1} _{E_{n,i}}}$

Show that ${\displaystyle \varphi }$ is simple and, except on F, ${\displaystyle |f-\varphi |<\varepsilon }$.

Finally, use all of the above (with the help of results proved in earlier exercises) to prove the desired result.