# If $f_n\to f$ in $L^1(\Bbb R)$ and $f_n\to g$ in $L^2(\Bbb R)$,deduce that $f=g$ a.e.

Let $$f_n$$ be a sequence in $$L^1(\Bbb R)\cap L^2(\Bbb R)$$ ;$$f\in L^{1}(\Bbb R)$$ and $$g\in L^2(\Bbb R)$$.

If $$f_n\xrightarrow {L_1(\Bbb R)} f, f_n\xrightarrow {L_2(\Bbb R)} g$$,then show that $$f=g$$(almost everywhere).

Attempt:

We have $$\int_\Bbb R |f_n-f|<\infty$$ and $$\int_\Bbb R |f_n-g|^2<\infty .$$

Now since $$f_n\xrightarrow {L_1(\Bbb R)}f\implies$$ there exists a subsequence $$f_{n_k}$$ of $$f_n$$ converging pointwise to $$f$$ a.e. $$\implies ||f_{n_k}-f||_{L_1(\Bbb R)}\to 0\implies \int _\Bbb R|f_{n_k}-f|\to 0$$.

Similarly there exists a subsequence $$f_{n_k}$$ of $$f_n$$ converging pointwise to $$g$$ a.e. $$\implies ||f_{n_k}-g||_{L_2(\Bbb R)}\to 0\implies \int _\Bbb R|f_{n_k}-g|^2\to 0$$.

Two questions:

• Is it true that $$\int _\Bbb R|f_{n_k}-g|^2\to 0\implies \int _\Bbb R|f_{n_k}-g|\to 0$$.

• If I can show that $$\int |f-g|=0$$ then that will imply that $$f=g$$ a.e.

• Do you know that when à sequence of functions converges in $L_1(A)$ (or $L_2(A)$) it is possible to find a subsequence that converges to same limit on A a.e. ? – user322559 Mar 13 '17 at 7:32
• The only reference I have is a PDF of the course I followed but it's in French. – user322559 Mar 13 '17 at 8:10
• @Ben: See the "properties" section here en.m.wikipedia.org/wiki/Convergence_in_measure – PhoemueX Mar 13 '17 at 8:43
• almost any book on measure theory has the a.e. subsequence result, for example Rudin RCA. – zhw. Mar 13 '17 at 19:56
• @Errol.Y;I just read about the subsequence criterion;How to proceed now – Learnmore Mar 16 '17 at 5:05

Because $f_n\to f$ in $L^1$, there is a subsequence converging to $f$ pointwise a.e.. Then that subsequence still converges in $L^2$ to $g$, so it has a further subsequence converging pointwise a.e. to $g.$ This subsubsequence now converges pointwise to both $f$ and to $g$ a.e. implying that $f=g$ a.e.
• No you cannot conclude from $\int|f_n-g|^2\to 0$ that $\int|f_n-g|\to 0$, and you cannot even conclude that $\int|f_n-g|$ is ever finite, because you do not no a priori that $g$ is in $L^1$. (Of course we know in this problem that it ends up being true because $g=f$.)
• If you could show that $\int|f-g|=0$, then it would follow that $f=g$ a.e., but I don't see a way to go this way, again especially because you do not know at first even that $g$ is in $L^1$. An $L^2$ limit of $L^1$ functions need not be in $L^1$. The conclusion of this problem does show that $g$ is in $L^1$, by showing that it equals $f$.