# $\int^1_0 \max\{f_1(x), . . . , f_n(x)\} \, dx \leq C$, $f_n \geq 0$ and $f_n \to 0$, then $f_n \to 0$ in $L^1(0,1)$

Let $f_n \in L^1(0, 1)$ and $C > 0$ be such that $f_n \geq 0$, $f_n \to 0$ a.e. in $(0, 1)$, and $$\int^1_0 \max\{f_1(x), . . . , f_n(x)\} \, dx \leq C$$ for every $n$. Prove that $f_n \to 0$ in $L^1(0,1)$.

My solution.

I define $g_n(x)=\max\{f_1(x), . . . , f_n(x)\}$. First of all I observe that $0\leq f_n \leq g_n$ a.e. The sequence $g_n$ is increasing and positive, then there exists a function $g\geq 0$ (eventually $g\equiv +\infty$) such that $g_n\to g$ a.e. By monotone convergence theorem we get $$\int^1_0 g\, dx = \lim_{n\to\infty}\, \int^1_0 g_n \, dx\leq C ,$$ then $g\in L^1(0,1)$ (and in particular a.e. finite). Now we are in position to use the "generalized" dominated convergence theorem (e.g. Variant of dominated convergence theorem, does it follow that $\int f_n \to \int f$?).

Do you agree with my proof ? I am also interested in different kind of solution.

Update: it is not necessary to use the generalized DCT. If fact, since $g_n$ is increasing, we have $0\leq f_n \leq g_n \leq g$. So we can use the classical DCT using $g$ as dominant function for the sequence $f_n$. My mistake.