Application of Egoroff's Theorem in $M$-measurable functions

Question

Here's the theorem that I'm trying to prove;

Here's how I have tried to solve it; Since $||f||_\infty$ = sup{ R > $0$:|$\mu$ ({x $\in$ X:| f(x) | > R) > $0$} ; $||f||_\infty$= sup |f(x)| so, $f$ is a positive measurable function i.e., $f\in$ $L^+$(X, $\mu$). So, we apply Egoroff's theorem here which states; "$f_n$ $\to$ $f$ $\mu$. a.e $\implies$ $f_n$ $\to$ $f$ almost uniformly."

Now, by the definition of Essential Uniform Convergence in $L^\infty$; " For all $\epsilon$>0, there exist E $\in$ $M$ : $\mu$(E) = $0$ and $f_n$ $\to$ $f$ on $E^c$, for all x$\in$X."

For this, let us have a decreasing sequence $E_k$= $\sum$$1/2^k$ and $X \supseteq E_1 \supseteq E_2 \supseteq E_3 \supseteq...$ and $E = \bigcap E_k$ so by continuity from above , $\mu (E_k)$ $\to 0$ as $k \to 0$ and $\mu(E_1) < \infty$.

Now, by the definition of Essential Uniform Convergence,

$lim ||f_n - f||_\infty$ $\to 0$ on $E_k^c$

Also, By a result, " $f_n \in L^+(X, \mu)$ $\to$ $f$ and $lim f_n=f$ a. e.Then, there exists a subsequence $f_{n_{k}}$ $\to$ $f_n$ $\mu$.a.e."

$\implies$ $lim || f_{n_{k}} - f||_\infty$ $\to$ $0$ on $E_k^c$

$\implies$ $lim || f_{n_{k}} - f||_\infty$ $=$ $0$ on $E_k^c$

$\implies$ $lim || $1_${E_k^c}$$ f_{n_{k}} - f||_\infty$ $=$ $0$ on $E_k^c$

However, my Professor says that I have correctly invoked Egoroff's theorem, but the subsequence is not chosen in conjunction with the descending sequence of sets. I need help in correcting my solution.

Answer 1

For each $k$, there is, by Egorov's theorem, a set $A_k\in\mathcal{M}$ such that $\mu(X\setminus A_k)<2^{-k}$, such that such that $f_n(x)\rightarrow f(x)$ uniformly for all $x\in A_k$. Thus, for some $n_k$, $\|(f-f_n)\mathbb{1}_{A_k}\|_\infty<2^{-k}$ for all $n\geq n_k$. One can choose the $n_k$'s so that $n_k<n_{k+1}$ (why?).

Let $E=\bigcap_k\bigcup_{j\geq k}(X\setminus A_j)$. Notice $$\mu(E)=\lim_k\mu(\bigcup_{j\geq k}X\setminus A_j\big)\leq\lim_k\sum_{j\geq k}\mu(X\setminus A_j)=\lim_k2^{-k+1}=0$$ Let $E_k=\bigcup_{j\geq k}(X\setminus A_j)$. Clearly $E_k$ is a decreasing sequence. Check that the sequence of $E_k$'s does the job.

Answer 2

Since $||f||_\infty$ = sup{ R > $0$:|$\mu$ ({x $\in$ X:| f(x) | > R) > $0$} ; $||f||_\infty$= sup |f(x)| so, $f$ is a positive measurable function i.e., $f\in$ $L^+$(X, $\mu$). So, we apply Egoroff's theorem here which states; "$f_n$ $\to$ $f$ $\mu$. a.e $\implies$ $f_n$ $\to$ $f$ almost uniformly."

Now, by the definition of Essential Uniform Convergence in $L^\infty$; " For all $\epsilon$>0, there exist E $\in$ $M$ : $\mu$(E) = $\epsilon$ and $f_n$ $\to$ $f$ on $E^c$, for all x$\in$X."

For this, By Egoroff's, let us have a decreasing sequence $\mu(E_k)$= $1/2^k$ and on $E^c$

$\implies \mu(E^c)= 1/2^k$ and $f_n \to f$ uniformly on $E^c$

In particular, there exists $n_k$ such that $n_K<n_{k+1}$

By induction, we define, $X \supseteq E_1 \supseteq E_2 \supseteq E_3 \supseteq... E_k$ with $\mu(E^c)= 1/2^k$ and $||(f_{n_k} - f)1_{E^c}||_\infty < 1/2^k$

Applying Egoroff's to $E^c$ to get some $F \subset E_k$ with $\mu(F) < 1/2^{k+1}$ such that $f_n$ $\to$ $f$ uniformly on $E^k/F$

Since $f_n$ $\to$ $f$ uniformly on $X/E^k$ ($E^k/F \bigcup X/E^k = X/F$

$\implies f_n$ $\to$ $f$ uniformly on $X/F$

Now, let $E_{k+1} = F$ and let $n_{k+1}>n_k$ such that $||(f_{n_{k+1}} - f)1_{E^{k+1}}||_\infty < 1/2^{k+1}$ as $k\to \infty$

So, we get the required result.