$f_{n}$ converges in measure to $f$ and $\{f_n\}$ is an increasing sequence . Prove $f_n$ converges to $f$ almost everywhere.

Question

Let $m$ be the Lebesgue measure. $E \subset \mathbb R$ is a measurable set. If $\{ f_n \}$ converges to $f$ in measure on $E$ and $\{f_n\}$ in increasing sequence then $f_n$ converges to $f$ almost everywhere.

Since $\{ f_n \}$ converges to $f$ in measure then $\forall \epsilon >0$ and $\forall \delta >0$ there is a $N \in \mathbb N$ such that $$ m(\{ x \in E : |f_n (x) - f(x)| \ge \epsilon \} ) < \delta $$

I have to prove that there is a $A \subset E $ such that $m(A)=0$ and $\forall x \in E \backslash A, f_n(x) $ converges to $f(x)$.

Let $$ A_n := \{ x \in E : |f_n (x) - f(x)| \ge \epsilon \} $$ Then $ \forall n \ge N $, $m(A_n) < \delta $

Let $A:= \bigcap_{ n \in \mathbb N} A_n$ then $$ m(A) \le m(A_N) < \delta$$ Then $\forall \delta >0, m(A) < \delta \Rightarrow m(A)=0$.

Let $x \in E \backslash A \Rightarrow \exists n_0 \in \mathbb N $ such that $x \notin A_{n_0} \Rightarrow |f_{n_0} (x) - f(x)| < \epsilon $

Since $\{f_n\}$ is an increasing sequence then $\forall n \in \mathbb N, f_n \le f_{n+1} $

If I can prove that $\forall n \in \mathbb N, f_n \le f $ Then $ \forall n \in \mathbb N$; $$ |f_n (x) - f(x)| = f (x) - f_n(x) \ge f (x) - f_{n+1}(x) = |f_{n+1} (x) - f(x)|$$ Then $\forall n \in \mathbb N, A_{n+1} \subset A_n $

So $\forall n \ge n_0, x \notin A_n \Rightarrow \forall n \ge n_0,|f_{n_0} (x) - f(x)| < \epsilon $.

But I couldn't prove that $\forall n \in \mathbb N, f_n \le f $. Also I can't find a way to prove that without using this condition.

Any help?

Answer 1

Suffices to assume that $m(E)<\infty$. Since $f_n$ converges to $f$ in measure in $E$, that is, for every $\varepsilon>0$, $\lim_n m(\{|f_n-f|>\varepsilon\}\cap E)=0$, there is a subsequence $n_k\nearrow\infty$ along which $f_n$ converges to $f$ on $E$ $m$-a.s. This is a well known fact (see for example this posting).

Say, $f_{n_k}$ converges pointwise to $f$ on $F\subset E$ where $m(E\subset F)=0$. For $x\in F$, the sequence $f_{n_k}(x)$ converges to $f(x)$. From this it is easy to check that in fact $f_n(x)$ converges to $f(x)$.

Here is what one can prove to justify the claim above:

If $a_n$ is monotone increasing sequence which converges along some subsequence to a number $a$, then the whole sequence $a_n$ converges to $a$.

Indeed, suppose $a_{n_k}$ is a convergent subsequence. Without loss go generality assume $n_0=1$. Now, as $a_n$ is monotone increasing, $a_{n_k}\leq a_{n_{k+1}}$ and so, $a=\lim_ka_{n_k}=\sup_ka_{n_k}<\infty$. Given $\varepsilon>0$, there is $k_0$ such that for $k\geq k_0$ $$ |a-a_{n_k}|=a-a_{n_k}<\varepsilon$$ If $n\geq n_{k_0}$ , then there is $k\geq k_0$ such that $n_k\leq n<n_{k+1}$. Thus $$a-\varepsilon<a_{n_k}\leq a_n\leq a_{n_{k+1}}\leq a$$ This shows that $a_n$ converges and that $\lim_na_n=a$.