# Let $f_n\to f$ $m$-almost everywhere, and that $|f_n|^2 dm \leq 1$ for each $n$. Prove that $f_n$ converges to $f$ in $L^1$.

Let $$f_n$$ be a sequence of Lebesgue measurable functions on the interval $$[0,1]$$. Assume that $$f_n$$ converges to a function $$f$$ $$m$$-almost everywhere, and that $$\int_{[0,1]}|f_n|^2 dm$$ for each $$n$$. Prove that $$f_n$$ converges to $$f$$ in $$L^1$$.

I'm supposed to use Egorov's theorem, but I don't see how that is useful here. Any hints would be appreciated!

• You mean $\int |f_n|^2 dm \leq 1$? – Teresa Lisbon Jan 17 at 3:10
• Yes, thank you, I fixed the typo above. – J. Bishop Jan 17 at 3:11
• What happens if you apply Egorov's theorem blindly to the above $f_n$? Just give yourself an idea of what to do first by blindly applying the theorem, then shaping the application to suit your result. Alternately, state Egorov's theorem so that my statement matches with yours. – Teresa Lisbon Jan 17 at 3:13

Using Egorov's Theorem, we can find for all $$\epsilon$$, a set $$B_{\epsilon}$$ of measure less then $$\epsilon$$ where the uniform convergence fails. Then,
$$\int|f-f_n|dm \leq \int_{B_{\epsilon}}|f-f_n|dm+\int_{B_{\epsilon}^c}|f-f_n|dm\leq \int 1_{B_{\epsilon}}|f-f_n|dm+\int_{B_{\epsilon}^c}|f-f_n|dm$$ The second term tends to zero as $$n$$ goes to infinity by uniform convergence so we need only control the first. Applying Cauchy Schwarz to the first term, $$(\int 1_{B_{\epsilon}}|f-f_n|dm)^2\leq \int 1_{B_{\epsilon}}dm\int|f-f_n|^2dm\leq \epsilon\int|f-f_n|^2dm\\$$
So, if you could show that $$\int|f-f_n|^2dm$$ is bounded uniformly for all $$n$$ you are finished. But, $$||f-f_n||_2\leq ||f||_2+||f_n||_2 \leq \liminf||f_n||_2+1\leq 2.$$
• Yes. By Fatou's Lemma, $\int |f|^2\leq \liminf\int |f_n|^2\leq 1$, so the last integral is bounded. – user254433 Jan 17 at 4:41