5
$\begingroup$

I'm dealing with this problem.

Let $(\Omega,\mathcal{F},\mu)$ be a measure space and $\{f_n\}$ a sequence of nonnegative integrable functions. Suppose $f_n\xrightarrow{\mu} f_0$ and $\displaystyle\int_\Omega f_nd\mu\xrightarrow{n\rightarrow\infty}\int_\Omega f_0d\mu<\infty$. Prove that $\displaystyle\int_{\Omega}|f_n-f_0|d\mu\xrightarrow{n\rightarrow\infty} 0$.

If the convergence of $\{f_n\}$s to $f_0$ were almost everywhere, then the problem would become Scheffé's lemma.

By convergence in measure, for each $\epsilon>0$ and natural $n$, define : $\displaystyle E^\epsilon_n:=\{\omega\in\Omega\;;\;|f_n-f_0|(\omega)\geq\epsilon\} \;\wedge\;F_n^\epsilon:=\Omega-E^\epsilon_n$ $$\int_\Omega |f_n-f_0|d\mu=I_1+I_2\quad;\quad I_1=\int_{E^\epsilon_n} |f_n-f_0|d\mu\;\wedge\;I_2=\int_{F^\epsilon_n} |f_n-f_0|d\mu$$ We have $\mu(E_n^\epsilon)\rightarrow 0$ and $|f_n-f_0|<\epsilon$ over $F_n^\epsilon$. But I can't control the terms $|f_n-f_0|$ in $I_1$ and $\mu(F_n^\epsilon)$ in $I_2$ when the measure of space is not finite !

EDIT

My book has never mentioned "Scheffé's lemma".

$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Let $\{f_{n_k}\}$ be a subsequence of $\{f_n\}$. Since $f_n \xrightarrow{\mu} f_0$, there exists a subsequence $\{f_{n_{k_j}}\}$ of $\{f_{n_k}\}$ such that $f_{n_{k_j}} \to f_0$ almost everywhere. Since $\int_\Omega f_{n_{k_j}}\, d\mu \to \int_\Omega f_0\, d\mu$, by Scheffé's lemma, $\int_\Omega |f_{n_{k_j}} - f_0|\, d\mu \to 0$. Since $\{f_{n_k}\}$ was arbitrary, $\int_\Omega |f_n - f_0|\, d\mu \to 0$.

Alternative: Since $f_n + |f_0| - |f_n - f_0| \ge 0$ and $\{f_n + |f_0| - |f_n - f_0|\}$ converges in measure to $2|f_0|$, Fatou's lemma gives

$$\int_\Omega 2|f_0|\, d\mu \le \varliminf \int_\Omega (f_n + |f_0| - |f_n - f_0|)\, d\mu = 2\int_\Omega |f_0|\, d\mu - \varlimsup \int_\Omega |f_n - f_0|\, d\mu.$$

Since $\int_\Omega |f_0|\, d\mu$ is finite (since $f_0 \ge 0$ a.e., so that $\int_\Omega |f_0|\, d\mu = \int_\Omega f_0 < \infty$), we deduce $\varlimsup \int_\Omega |f_n - f_0|\, d\mu \le 0$. Hence $\int_\Omega |f_n - f_0|\, d\mu \to 0$.

$\endgroup$
14
  • 1
    $\begingroup$ The point is, a sequence $\{x_n\}$ of real numbers converges to a point $x$ if, and only if, every subsequence of $\{x_n\}$ has a subsequence which converges to $x$. In the case of the problem, $x_n = \int_\Omega |f_n - f_0|$. $\endgroup$
    – kobe
    Apr 9, 2015 at 17:56
  • 1
    $\begingroup$ Why didn't you mention this in the body of your question? It seemed like you knew Scheffé's lemma, but didn't see how it could be applied to the problem. In any case, I made an edit and gave an alternative proof. $\endgroup$
    – kobe
    Apr 9, 2015 at 18:47
  • 1
    $\begingroup$ I was supposed to put $|f_0|$ in place of $f_0$, but note $f_0 \ge 0$ a.e., so $\int_\Omega |f_0|\, d\mu = \int_\Omega f_0\, d\mu < \infty$. To see that $f_0 \ge$ a.e., note $f_n \xrightarrow{\mu} f_0$ implies $|f_n| \xrightarrow{\mu} |f_0|$. Since $f_n$ is nonnegative, $f_n \xrightarrow{\mu} |f_0|$. Since $f_n$ converges in $f_0$ and $|f_0|$ in $\mu$-measure, $f_0 = |f_0|$ a.e., i.e., $f_0 \ge 0$ a.e.. $\endgroup$
    – kobe
    Apr 9, 2015 at 23:39
  • 2
    $\begingroup$ @FardadPouran why do refer back to that version of Fatou's lemma when the one I used is for convergence in measure? Namely, if you have a sequence $(g_n)$ of nonnegative measurable functions on a measure space $(X,\mathcal{M},\mu)$ and $g$ is a measurable function such that $g_n$ converges in $\mu$-measure to $g$, then $\int_X g\, d\mu \le \varliminf \int_X g_n\, d\mu$. $\endgroup$
    – kobe
    May 7, 2015 at 14:29
  • 1
    $\begingroup$ Hi @D.R., it follows from the fact that a sequence converges if and only if every subsequence of that sequence has a further subsequence which converges to the same limit. $\endgroup$
    – kobe
    Nov 6, 2019 at 18:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .