Let $(\mathbb{R},\mathcal{L},m)$
Can someone explain to me why $f_{n}(x)=\chi_{(0,\frac{1}{n}]}$ converges almost everywhere to $0$ but not uniformly…
also why does $f_{n}(x)=n^{-1}\chi_{(0,n)}$ converge uniformly on $\mathbb{R}$?
[Math] almost everywhere convergence vs uniform convergence
measure-theoryreal-analysis
Related Solutions
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:
- $\int g_n\,d\mu=\int f_n\,d\mu$ for all $n$, since these functions differ on a set of measure $0$; similarly, $\int g\,d\mu=\int f\,d\mu$.
- 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 $\chi_{[2,3]}(1.5)=0$.
Actually if $(f_n)_{n \in \mathbf{N}}$ is a sequence of measurable functions to a complete space, then the set of convergence points of $(f_n)$ is always measurable. In your example, $f_n : D \to \mathbf{R}$ and $\mathbf{R}$ is a complete space.
$$\{x \in D \; \vert \; (f_n(x)) \text{ converges}\}=\bigcap_{\epsilon \in \mathbf{Q_+^*}}\bigcup_{N \in \mathbf{N}}\bigcap_{n,m\ge N}\{x \in D \; \vert \; |f_n(x)-f_m(x)| \le \epsilon\}$$
And $\{x \in D \; \vert \; |f_n(x)-f_m(x)| \le \epsilon\}=|f_n-f_m|^{-1}([0,\epsilon])$ is measurable because $|f_n-f_m|$ is still measurable.
Best Answer
(1) For any $x_0\in \mathbb{R}$, if $x_0\not\in (0,1]$ then $f_n(x_0)=0$ for all $n$; if $x_0\in (0,1]$ there exists $N$ such that when $n>N$ we have $x_0\not\in (0,\frac{1}{n}]$, and then $f_n(x_0)=0$. So $\lim_{n\to \infty}f_n(x)=0$ for all $x\in \mathbb{R}$. (2) For $\epsilon_0=\frac{1}{2}$, for any $N\in \mathbb{N}$, we have that $\frac{1}{N+2}\in (0, \frac{1}{N+1}]$, hence $|\chi_{(0,\frac{1}{N+1}]}(\frac{1}{N+2})-0|=1\geq \epsilon_0$. This means that $\chi_{(0,\frac{1}{n}]}$ does not uniformly converge t0 $0$. (3) Since $|n^{-1}\chi_{(0,n)}|\leq \frac{1}{n}$ for all $x\in \mathbb{R}$. Then $\frac{1}{n}\chi_{(0,n)}$ uniformly converges to $0$.