Real Analysis – Monotone Convergence Theorem and Almost Everywhere Concept

measure-theoryreal-analysis

Could someone enlighten me. Monotone convergence theorem says: if $f_{n},f\in L^{+}$ i.e measurable functions, such that $f_{n}\uparrow f$ pointwise then $\int f_{n}\uparrow \int f$.

But then my teacher says we can relax the condition, using almost everywhere.

We can assume that $f_{n}\uparrow f$ a.e $\Rightarrow \int f_{n}\uparrow \int f$ and then he does something I don't get.

He asks the question:

''Is it true $f_{n}\to f$ a.e $\Rightarrow \int f_{n}\to \int f$?''

what does the horizontal arrow mean in this case? (does he also mean converge to but almost everywhere?)

Here is his counterexample
$(X,\mathcal{M},\mu)=(\mathbb{R},\mathcal{L},m)$ and we let $f_{n}=\chi_{[n,n+1]}$

then $\int \chi_{[n,n+1]}dm=m([n,n+1])=1$ but $f_{n}\to 0$ pointwise

Is the horizontal and the vertical arrow meaning the same (''converge to'')?

Why do the characteristic functions converge pointwise??

Best Answer

First, for the "almost-everywhere" MCT:

Suppose that $f_n,\ f:X\rightarrow\mathbb{R}$ are such that $f_n\nearrow f$ a.e., where $(X,\mathcal{A},\mu)$ is a measure space.

This means that there is a measureable set $E\subseteq X$ such that $\mu(E)=0$, and we have that $f_n(x)\nearrow f(x)$ for all $x\in X\setminus E$.

If we define new functions $g_n,g:X\rightarrow\mathbb{R}$ by $$ g_n(x)=\begin{cases}f_n(x) & \text{if }x\notin E\\0 & \text{if }x\in E\end{cases},\qquad g(x)=\begin{cases}f(x) & \text{if }x\notin E\\0 & \text{if }x\in E\end{cases}. $$ Why should we care about these functions? You can prove:

  1. $\int g_n\,d\mu=\int f_n\,d\mu$ for all $n$, since these functions differ on a set of measure $0$; similarly, $\int g\,d\mu=\int f\,d\mu$.
  2. For every $x\in X$ (not just $x\in E$), $g_n(x)\nearrow g(x)$. Thus, by the MCT, $\int g_n\,d\mu\rightarrow\int g\,d\mu$.

Combining these, we see that $\int f_n\,d\mu\rightarrow\int f\,d\mu$, even though we relaxed our monotonicity assumption a bit.

Now, for his counterexample if we completely remove the monotonicity condition: here we have $f_n:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f_n=\chi_{[n,n+1]}$.

You asked why $f_n$ converges pointwise? Pick any $x\in \mathbb{R}$. Note that if we let $n\rightarrow\infty$, then eventually $n>x$; but if $n>x$, then $\chi_{[n,n+1]}(x)=0$. This holds for all $n>x$, so that $\chi_{[n,n+1]}(x)\rightarrow 0$ as $n\rightarrow\infty$.

This holds for every $x\in\mathbb{R}$; so, the sequence $\{\chi_{[n,n+1]}\}$ converges pointwise to the $0$-function.

This stands in contrast to the fact that, as your instructor said, $\int\chi_{[n,n+1]}\,d\mu=\mu([n,n+1])=1$ when we take $\mu$ to be Lebesgue measure.

Why does this not violate the MCT? Because the sequence isn't monotone increasing! Take, for instance, the fact that $\chi_{[1,2]}(1.5)=1$, but $\chi_{[2,3]}(1.5)=0$.

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