Bounded sequence in $L^\infty$ which converges in $L^1$

Question

I have sequence in $L^\infty(\mathbb{R}^n)\cap L^1(\mathbb{R}^n)$ such that
1. $(u_n)_n$ is bounded in $L^\infty$ : there exists $a>0$ such that $\|u_n\|_{L^\infty}\leq a$.
2.$(u_n)_n$ converges (strongly) in $L^1$ : there exists $u\in L^1(\mathbb{R}^n)$ such that $\|u_n-u\|_{L^1}\to0$.
Is it true that the limit $u$ actually belongs to $L^\infty(\mathbb{R}^n)$ and that $\|u\|_{L^\infty}\leq a$ ? I guess that a strating point could be to take a weak-* converging subsequence in $L^\infty$ and then prove that this limit is $u$...

Answer 1

For a more direct proof, let $\lambda$ denote the Lebesgue measure and $\delta>0$. Using Chebychev's inequality we get $$\lambda\left(\left\{x: \lvert u(x) \rvert \geq a+\delta \right\}\right) \leq \lambda\left(\left\{x: \lvert u(x) -u_{n}(x) \rvert \geq \delta/2 \right\} \right) + \underbrace{\lambda \left(x: \lvert u_{n}(x)\rvert \geq a + \delta/2 \right)}_{=0} \leq\\ \frac{2}{\delta} \, \lvert\lvert u-u_{n}\rvert\rvert_{L^{1}}\rightarrow 0$$ as $n\rightarrow \infty$. Since $\delta>0$ was arbitrary, we arrive at the conclusion that $u$ is essentially bounded in $\mathbb{R}^{n}$ with norm less than $a$.

Answer 2

Since $u_n\to u$ in $L^1,$ there is a subsequence $u_{n_k}$ that converges to $u$ a.e. We have each $|u_{n_k}|\le a$ a.e., so $|u|\le a$ a.e., giving the result.