# On convergence a.e and convergence measure

I have a question. First, I know that convergence in measure of a sequence of functions $f_n$ is different than convergence a.e., wich means there are sequences that converge in measure but not a.e. but this excercise got me in doubt if there is some kind of duality between convergence in measure and convergence a.e.

Excercise: Let $f, f_n \in L_1$, and suppose that $f_n \rightarrow f$ a.e. Show that $||f_n -f||_1 \rightarrow 0$ if and only if $||f_n||_1 \rightarrow ||f||_1$ (Note that this result also hold if "a.e." is replaced by "in measure")

My attempt at Proof:

$\big($$\Leftarrow : If ||f_n||_1 \rightarrow ||f||_1 and f_n \rightarrow f a.e. then,$$\lim\int |f_n -f| \leq \lim \int |f_n| + |f| = 2\int|f|$$Now applying Fatou's Lemma:$$0 = \int \lim|f_n - f| = \int \underline{\lim}|f_n - f| \leq \underline{\lim}\int |f_n - f|\leq \overline{lim} \int |f_n - f| \leq \int \overline{\lim}|f_n - f| = 0$$\big($$\Rightarrow$ : $$0 \leq \lim \int ||f_n| - |f|| \leq \lim\int |f_n - f| \leq \int \lim |f_n - f|= 0$$ Wich completes the proof for $f_n \rightarrow f$ a.e.

Now, and this part I'm not that sure, if $f_n \rightarrow f$ in measure, suppose $\lim \int |f_n - f| \neq 0$, that is there is some $\delta > 0$ and a subsequence $f_{n_k}$ such that $\lim \int |f_{n_k} - f| \geq \delta$. But we know that every sequence wich converges in measure to some $f$ has a subsequence that converges a.e., in this case there exists a sub-subsequences $f_{n_{k_j}} \rightarrow f$ a.e. But now $0 = \lim \int |f_{n_{k_j}} - f| \geq \delta$ wich is a contradiction.

The question is: Can we generalize the process above?

For example is P is property on $f_n$ a sequence of functions, can we assure that a result will hold if we replace "a.e." by "in measure" in general? Can you provide a counter-example to 2)?

1. If $f_n \rightarrow f$ in measure and P is valid for $f_n$, then if $f_n \rightarrow f$ a.e. it would mean $f_n \rightarrow f$ in measure too so P would still be valid.

2. But if $f_n \rightarrow f$ a.e. and P is valid for the sequence, can we arrive that is valid if we replace convergence a.e. by convergence in measure?

• How do you pass from $f_n\to f$ a.e. to $\int |f_n-f|dx\to0$? – Milly Oct 20 '14 at 4:25
• @Milly I proved that in the first part. – Aram Oct 20 '14 at 4:34
• Which conditions do you assume in your question? convergence a.e. and convergence in measure are unrelated in general. Your assumption $\|f_n\|_{L^1}\to \|f\|_{L^1}$ is quite strong, do you assume it in your question? – Milly Oct 20 '14 at 4:46
• @Milly No, I dont. – Aram Oct 20 '14 at 4:53
• Then you cannot pass from $f_n\to f$ a.e. (nor from $f_n\to f$ in measure) to $\int |f_n-f|dx\to 0$ (e.g. $f_n=n\chi_{[0,1/n]}$ converges to $0$ a.e. and in measure, but not in $L^1$). – Milly Oct 20 '14 at 4:57