How to prove convergence of integrals using almost everywhere convergence?

[Chris L T521]

Here's this week's problem.

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Problem: Suppose that $\{f_n(x)\}$ and $\{g_n(x)\}$ are two sequences of measurable functions (on $\mathbb{R}$) such that $|f_n(x)|\leq g_n(x)$ for each $n=1,2,\ldots$. And suppose that $f_n$ converges to $f$ and that $g_n$ converges to $g$ almost everywhere. Show that
\[\lim_{n\to\infty}\int_{\mathbb{R}} g_n\,dm = \int_{\mathbb{R}} g\,dm\]
implies that
\[\lim_{n\to\infty}\int_{\mathbb{R}} f_n\,dm = \int_{\mathbb{R}} f\,dm\]

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[Chris L T521]

This week's problem was correctly answered by PaulRS and girdav.

Here's my solution:

Spoiler

Proof: Let $\{g_n\}$ be a sequence of Lebesgue integrable functions such that $g_n\rightarrow g$ almost everywhere and $\displaystyle\int_{\mathbb{R}} g_n\,dm\rightarrow \int_{\mathbb{R}} g\,dm$, and let $\{f_n\}$ be a sequence of Lebesgue measurable functions such that $|f_n|\leq g_n$ and $f_n\rightarrow f$ almost everywhere. Since $|f_n|\leq g_n$, we have $|f|\leq g$. Thus, $|f_n-f|\leq |f_n| + |f|\leq g_n+g$ and $\{g_n+g-|f_n-f|\}$ is a sequence of non-negative measurable functions. Therefore, by Fatou's lemma, we have\[\int_{\mathbb{R}}\liminf(g_n+g-|f_n-f|)\,dm\leq \liminf\int_{\mathbb{R}} (g_n+g-|f_n-f|)\,dm\]
This implies that
\[\int_{\mathbb{R}} 2g\,dm \leq \int_{\mathbb{R}} 2g\,dm+\liminf\left(-\int_{\mathbb{R}}|f_n-f|\,dm\right)=\int_{\mathbb{R}} 2g\,dm -\limsup\int_{\mathbb{R}} |f_n-f|\,dm\]
Hence, we see that
\[\limsup\int_{\mathbb{R}}|f_n-f|\,dm\leq 0\leq \liminf\int_{\mathbb{R}}|f_n-f|\,dm\]
and thus we have $\displaystyle\int_{\mathbb{R}}|f_n-f|\,dm\rightarrow 0$.Therefore, we see that $\displaystyle\int_{\mathbb{R}} g_n\,dm\rightarrow \int_{\mathbb{R}} g\,dm$ implies that $\displaystyle\int_{\mathbb{R}} f_n\,dm\rightarrow\int_{\mathbb{R}} f\,dm$. Q.E.D.

and here's PaulRS's solution:

Spoiler

Since $|f_n| \leq g_n$ we have that $g_n - f_n \geq 0$ and $g_n + f_n \geq 0$, and so we have $L^+$ functions and we may apply Fatou's Lemma:

\[\int g dm - \int f dm = \int (g-f)dm = \int \liminf (g_n - f_n) dm \leq \liminf \int (g_n - f_n) dm = \liminf \left( \int g_n dm - \int f_n dm \right) = \int g dm - \limsup \int f_n dm \]where we have used that $\liminf g_n = g$ and $\liminf f_n = f$ almost everywhere, $\lim \int g_n dm = \int g dm$ (and so the liminf is the sum of the limit plus the liminf) and $-\limsup \left( ... \right) = \liminf (-...)$.Similarly : \[\int g dm + \int f dm = \int (g+f)dm = \int \liminf (g_n +f_n) dm \leq \liminf \int (g_n+ f_n) dm = \liminf \left( \int g_n dm + \int f_n dm \right) = \int g dm + \liminf \int f_n dm \]Now, these two inequalities translate to\[\limsup \int f_n dm \leq \int f dm \leq \liminf \int f_n dm\]which proves the assertion. $\square$