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 $L^1$.


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.


1 Answer 1


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 $n$.

  • $\begingroup$ Thanks a lot. It helped. $\endgroup$
    – player1235
    Dec 22, 2020 at 15:08
  • 1
    $\begingroup$ you're welcome. in this case, you might consider upvoting or accepting the answer $\endgroup$
    – Targon
    Dec 22, 2020 at 15:54
  • $\begingroup$ did you use Cauchy Schwartz on the inequalities after "therefore"? To use $\int |f_n|^2\leq 1$ @Targon $\endgroup$ Jun 8, 2021 at 3:05
  • $\begingroup$ I'll assume you meant the first 'Therefore'. In the following line I used the triangle inequality for the first $\leq$ sign, i.e. $\vert f_n - f \vert \le \vert f_n \vert + \vert f \vert$. $\endgroup$
    – Targon
    Jun 9, 2021 at 16:58

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