Let $\Omega \subseteq \Bbb R^d$ be a bounded and measurable set and let $f, f_n : \Omega \to \Bbb R\cup \{±∞\}$ for $n ∈ \Bbb N$ be measurable such that :
- $f, f_n ∈ L^1(\Omega)$ for all $n ∈ N$ ,
- $\lim_{n \to \infty} \lVert f_n − f \rVert _{L^\infty(\Omega)} = 0$
Prove that $$ \lim_{n \to \infty} \int_\Omega f_n = \int_\Omega f$$
I was told to use the dominated convergence theorem but I used a simpler approach and was wondering if it was correct : since $f, f_n ∈ L^1(\Omega)$, then $\lvert \int_\Omega f_n - \int_\Omega f \rvert < \infty$ by triangular inequality. Then, since $\Omega$ is bounded and measurable, it has finite measure $m(\Omega)\neq 0$ (otherwise the two integrals are already equal to $0$). By assumption 2, we have that $\forall \epsilon > 0, \exists N \in \Bbb N$ such that $\forall n > N$, $$\sup_{x \in \Omega}\lvert f_n(x) - f(x) \rvert < \frac{\epsilon}{m(\Omega)} $$ Finally, we can observe that $\forall n> N$, $\lvert \int_\Omega f_n - \int_\Omega f \rvert \leq \int_\Omega \lvert f_n-f\rvert \leq m(\Omega)\frac{\epsilon}{m(\Omega)} = \epsilon$.
Is it correct ? I still would like to know how could the monotone convergence work here, does it make things easier ?