1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Both implies convergence in measure and limit is unique? $\endgroup$ – Lost1 Jan 5 '14 at 16:32
  • 4
    $\begingroup$ Your solution looks good to me! $\endgroup$ – user940 Jan 5 '14 at 17:49
1
$\begingroup$

$L^1$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.