# Convergence in $L^1$ coincides with convergence a.e.

On the measure space $(X,F,m)$, if $\lim\limits_{n\rightarrow \infty} \int\limits_X |f_n-f|\mathrm{d}m=0$ and $\lim\limits_{n\rightarrow \infty} f_n =g$ almost everywhere, then prove that $f=g$ almost everywhere.

Well my proof is:

since $\lim\limits_{n\rightarrow \infty}f_n=g$, then $\liminf|f-f_n|=\lim\limits_{n\rightarrow \infty}|f-f_n|=|f-g|$ so $$0\leq\int_X|f-g|\mathrm{d}m \leq \liminf\int_X|f-f_n|\mathrm{d}m \text{ (by Fatou's lemma)}$$ $$=\lim_{n\rightarrow \infty} \int_X |f-f_n|\mathrm{d}m = 0$$

and therefore we must have $|f-g|=0$ a.e., or $f=g$ a.e. as required. Is this correct? It seems too easy to me and I'm afraid I might have done a mistake somewhere.

• Both implies convergence in measure and limit is unique? – Lost1 Jan 5 '14 at 16:32
• Your solution looks good to me! – user940 Jan 5 '14 at 17:49

$L^1$ convergence implies the existence of a sub-sequence of $(f_n)_{n\geq1}$ that converges pointwise a.e. to $f$. That sub-sequence must also converge to $g$ pointwise a.e. (as its parent sequence, $(f_n)_{n\geq1}$, converges to $g$ pointwise a.e.). Hence, $f$ and $g$ are equal a.e.