# Weak convergence and convergence in measure

Let $$\Omega \subset \mathbb{R}^n$$ be an open set and let $$1 < p < \infty$$. Consider a sequence $$(f_n)_n \subset L^p(\Omega)$$ and $$f \in L^p(\Omega)$$ such that $$f_n$$ converges to $$f$$ weakly in $$L^p(\Omega)$$. Let $$g_n: \Omega \rightarrow \mathbb{R}$$ and $$g: \Omega \rightarrow \mathbb{R}$$ be measurable functions such that $$g_n \rightarrow g$$ in measure and $$||g_n||_{L^\infty} \le C$$, $$\forall n$$ and for some constant $$C > 0$$. Prove that $$g \in L^\infty (\Omega)$$ and $$f_n g_n \rightharpoonup fg$$ in $$L^p(\Omega)$$.

$$\textbf{Note}$$: $$g_n \rightarrow g$$ in measure if $$\forall \epsilon > 0$$, $$\mu\{x \in \Omega : |g_n(x) - g(x)| > \epsilon\} \rightarrow 0$$, where $$\mu$$ in this case denotes Lebesgue measure.

Here is my attempt.

Since $$g_n \rightarrow g$$ in measure, there exists a subsequence $$(g_{n_k})_k$$ such that $$g_{n_k} \rightarrow g$$ a.e. in $$\Omega$$. It follows that $$|g_{n_k}(x)| \rightarrow |g(x)|$$ for a.e. $$x \in \Omega$$. By hypothesis, $$|g_{n_k}(x)|\le ||g_{n_k}||_{L^{\infty}} \le C$$ for all $$k \in \mathbb{N}$$ $$\Rightarrow$$ $$|g(x)| \le C$$ for a.e. $$x \in \Omega$$. This implies that $$g \in L^{\infty}(\Omega)$$.

We aim to prove that $$f_n g_n \rightharpoonup fg$$ in $$L^p(\Omega)$$ $$\Longleftrightarrow$$ $$\forall \phi \in L^q(\Omega)$$ (where $$\frac{1}{p} + \frac{1}{q} = 1$$) , $$\int_{\Omega} f_n g_n \phi dx \rightarrow \int_{\Omega} fg \phi dx$$. First of all, let us check $$f_n g_n \in L^p(\Omega)$$.

$$||f_n g_n||_{L^p}^p = \int_{\Omega} |f_n g_n|^p dx \le ||g_n||_{L^{\infty}}^p ||f_n||_{L^p}^p < \infty$$ since, by hypothesis, $$f_n \in L^p(\Omega), g_n \in L^\infty(\Omega), \, \forall n$$. (The same argument holds true also for $$fg$$).

Note that $$f_n g_n - fg = (f_n -f)g_n + f (g_n - g)$$. Then

$$|\int_{\Omega} f_n g_n \phi dx - \int_{\Omega} fg \phi dx| = |\int_{\Omega} (f_n g_n -fg) \phi dx| \le |\int_{\Omega} (f_n-f) g_n \phi dx| + |\int_{\Omega} f(g_n -g) \phi dx|$$.

Let us focus on the first integral on the right hand-side of the expression above. $$|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx| \le C |\int_{\Omega} (f_n-f) \phi dx| \rightarrow 0$$, since $$f_n \rightharpoonup f$$.

Let $$\epsilon > 0$$ and define $$A_{\epsilon} := \{x \in \Omega: |g_n(x) - g(x)| < \epsilon\}$$. Convergence in measure implies that $$|\int_{\Omega} f(g_n -g) \phi dx| \le |\int_{A_{\epsilon}} f(g_n - g) \phi dx| + |\int_{\Omega \setminus A_{\epsilon}} f(g_n -g) \phi dx| < \epsilon \int_{A_{\epsilon}} |f| \phi dx + \int_{\Omega \setminus A_{\epsilon}} |f(g_n -g) \phi| dx$$ $$< \epsilon \int_{\Omega} |f| \phi dx + \int_{\Omega \setminus A_{\epsilon}} |f(g_n -g) \phi| dx$$. The second integral tends to zero for $$n \rightarrow \infty$$ because $$g_n \rightarrow g$$ in measure (which means that $$\mu(\Omega \setminus A_{\epsilon}) \rightarrow 0$$, the thesis is a consequence of the absolute continuity of the integral), while $$\int_{\Omega} |f| \phi dx$$ is merely a constant.

If anyone could check the above reasoning, it would be greatly appreciated.

• Everything looks good except for one small detail: to conclude that $\int_{\Omega\setminus A_\epsilon} |f(g_n-g)\phi| \to 0$ you need an integrand which does not depend on $n$. You can do so by first using the hypothesis $\lVert g_n \rVert_{L^\infty}\leq C$: $$\int_{\Omega\setminus A_\epsilon} |f(g_n-g)\phi| \leq \lVert g_n - g\rVert_{L^\infty} \int_{\Omega\setminus A_\epsilon} |f\phi| \leq 2C\int_{\Omega\setminus A_\epsilon} |f\phi|,$$ and now you can use continuity of the integral. Jan 15, 2022 at 20:25
• Thank you very much! Jan 15, 2022 at 20:50
• @Hilbert1234 I don't think $|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx|$ is correct .. Usually we don't have $|\int fg| \le |g|_\infty|\int f|$ ..
– r9m
Jan 16, 2022 at 13:35
• I am using the fact that, by definition of $||\cdot||_{L^{\infty}}$, $g_n(x) \le ||g_n||_{L^{\infty}}$ a.e. in $\Omega$. Honestly, I don't see why it is not correct. Thanks. Jan 17, 2022 at 11:00
• And by hypothesis, $||g_n||_{L^{\infty}} \le C$, so I could have used straightforwardly $C$ instead of the $L^{\infty}$ norm. Jan 17, 2022 at 11:17

As pointed out by r9m, the step $$|\int_{\Omega} (f_n-f) g_n \phi dx| \le ||g_n||_{L^{\infty}} |\int_{\Omega} (f_n-f) \phi dx|$$ may not be valid, for example, if $$f=0$$, $$f_n=\mathbf{1}_{[n,n+1)}-\mathbf{1}_{(-n-1,-n]}$$ and $$g_n=\frac{1}{n}f_n$$, the right hand side is zero if $$\phi$$ is even.
It is actually better to slip in the following way: $$\int f_ng_n\phi-\int fg\phi=\int f_n\left(g_n-g\right)\phi+\int f_ng\phi-\int fg\phi$$ because one can directly exploit the weak convergence assumption to get the convergence to $$0$$ of the difference of the last two integrals. It remains to check that $$a_n:=\int f_n\left(g_n-g\right)\phi\to 0$$. To do so, define for a fixed $$\varepsilon$$ and a fixed $$n$$ the set $$A_{n,\varepsilon}=\{\lvert g_n-g\rvert>\varepsilon\}$$. Then $$\lvert a_n\rvert\leqslant \left\lvert \int_{A_{n,\varepsilon}}f_n\left(g_n-g\right)\phi \right\rvert+\left\lvert \int_{A_{n,\varepsilon}^c}f_n\left(g_n-g\right)\phi \right\rvert\leqslant 2C\int_{A_{n,\varepsilon}}\lvert f_n\rvert\lvert \phi\rvert +\varepsilon\int_{A_{n,\varepsilon}^c}\lvert f_n\rvert\lvert \phi\rvert$$ and by Hölder's inequality, we get $$\lvert a_n\rvert\leqslant 2C\sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi\mathbf{1}_{A_{n,\varepsilon}}\rVert_q+\varepsilon \sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi \rVert_q.$$ Since $$\mu(A_{n,\varepsilon})\to 0$$ for each positive $$\varepsilon$$, it follows that $$\lVert \phi\mathbf{1}_{A_{n,\varepsilon}}\rVert_q\to 0$$ hence $$\limsup_n \lvert a_n\rvert\leqslant \varepsilon \sup_{N\geqslant 1}\lVert f_N\rVert_p\lVert \phi \rVert_q$$, giving the result.