$f_n \rightarrow f$ in $L^1$ if $\mu(X) < \infty$, $f_n \rightarrow f$ almost uniformly and $\int_X |f_n|^2 d \mu \leq 1$

Suppose $$(f_n)$$ is a sequence of measurable functions on measure space $$(X, \mathcal{A}, \mu)$$. Suppose that $$\mu(X) < \infty$$, $$f_n$$ converges to $$f$$ almost everywhere and $$\int_X |f_n|^2 d \mu \leq 1$$ for all $$n \in \mathbb{N}$$. Prove that $$f_n$$ converges to $$f$$ in $$L^1$$.

Attempt:

By Egoroff's theorem, $$f_n$$ converges to $$f$$ almost uniformly (i.e. for all $$\epsilon >0$$, there exists $$E \subseteq X$$ such that $$\mu(E) < \epsilon$$ and $$f_n$$ converges to $$f$$ uniformly on $$E^c$$.)

\begin{align*} \lim_{n \rightarrow \infty} \|f_n-f\|_1 &= \lim_{n \rightarrow \infty} \int_X |f_n-f| d \mu \\ &= \lim_{n \rightarrow \infty} \left( \int_E |f_n-f| d \mu + \int_{E^c} |f_n-f| d \mu \right) \\ &= \lim_{n \rightarrow \infty} \int_E |f_n-f| d \mu + \lim_{n \rightarrow \infty} \int_{E^c} |f_n-f| d \mu \\ \end{align*}

Notice \begin{align*} \lim_{n \rightarrow \infty} \int_{E^c} |f_n-f| d \mu &\leq \lim_{n \rightarrow \infty} \int_{E^c} \sup_{E^c} |f_n-f| d \mu \\ &= \lim_{n \rightarrow \infty} \sup_{E^c} |f_n-f| \int_{E^c} d \mu \\ &= \left( \lim_{n \rightarrow \infty} \sup_{E^c} |f_n-f| \right) \mu(E^c) \\ &= 0 \end{align*} by uniform convergence of $$f_n$$ to $$f$$ on $$E^c$$.

My professor said to use Egoroff's theorem to do this proof but I am not sure how to proceed.

So because of $$f_n \in L^2$$ and $$\mu (E ) < \infty$$ we have the existence of some $$C>0$$ such that $$\int \vert f_n \vert ~ \mathrm d \mu \leq C$$ (so it is still uniformly bounded, in other words). Therefore $$\int_E \vert f_n - f \vert ~ \mathrm d \mu \leq \int_E \vert f_n \vert ~ \mathrm d \mu + \int_E \vert f \vert ~ \mathrm d \mu \leq \mu (E) \cdot \big(C + 1\big) \quad\text{for any } n \in \mathbb N .$$ With $$\mu (E) \to 0$$ we are done.
Note that you run into problems when handling the other integral like that, below 'Notice'. Because although $$f_n \to f$$ almost everywhere it could still be, that $$\sup \vert f_n - f \vert = \infty$$ for every $$n$$. So I would just use, that by uniform convergence, given an $$\varepsilon >0$$ we have $$\vert f_n - f \vert < \frac \varepsilon{\mu (E)}$$ on $$E^\mathrm c$$ and therefore $$\int_{E^\mathrm c} \vert f_n - f \vert ~ \mathrm d \mu < \varepsilon$$ for large enough $$n$$.
• did you use Cauchy Schwartz on the inequalities after "therefore"? To use $\int |f_n|^2\leq 1$ @Targon Jun 8, 2021 at 3:05
• I'll assume you meant the first 'Therefore'. In the following line I used the triangle inequality for the first $\leq$ sign, i.e. $\vert f_n - f \vert \le \vert f_n \vert + \vert f \vert$. Jun 9, 2021 at 16:58