Let $(X,S,\mu)$ be a measure space and let each $f_k:X \rightarrow [0, \infty)$ be $S$-measurable. I want to prove that if $lim_{k \rightarrow \infty} \int f_k d\mu = 0$ then $lim_{k \rightarrow \infty} f_k(x) = 0$ for almost every $x \in X$.
Let me first say that I am not even sure if the statement is true. Here is my attempt:
Let $a_k = \int f_k d\mu$, so $lim_{k \rightarrow \infty} a_k = 0$.
Let $\beta \in \mathbb{R}^+$; let $A_{k,\beta}$ be $\{x \in X:f_k(x) \ge \beta \}$; it must be the case that $\mu(A_{k,\beta}) < \frac{a_k}{\beta} $ otherwise $ \int f_k d\mu \ge a_k$.
Letting $k \rightarrow \infty$ in $\mu(A_{k,\beta}) < \frac{a_k}{\beta} $ gives $\lim_{k \rightarrow \infty}\mu(A_{k,\beta}) = 0$.
Assuming $f_k$ converges to a function $f$, $\lim_{k \rightarrow \infty}\mu(A_{k,\beta}) = 0$ implies that $f(x) < \beta + \epsilon $ for any $\epsilon > 0$ and for a.e. $x \in X$ because if this was not true then there would be a set of $B$ of finite and positive measure such that $f(x) \ge \beta + \epsilon$ for all $x \in B$ and by Egorov's theorem $B$ has a subset $B'$ such that $f_1,f_2,...$ converges uniformly to $f$ on $B'$ and $\mu(B') = \mu(B) - \delta$ (for any $\delta > 0$), this contradicts $\lim_{k \rightarrow \infty}\mu(A_{k,\beta}) = 0$. Hence if $f_k$ converges to a function $f$, $\lim_{k \rightarrow \infty}\mu(A_{k,\beta}) = 0$ implies that $f(x) \le \beta $ for a.e. $x \in X$.
Since $\beta$ is arbitrary, if $f_1, f_2,...$ converges at all then its limit $f$ must be $0$ almost everywhere in $X$.
How do I prove that $f_1, f_2,...$ is convergent? Is all of the above correct? Is there a simpler way (I am allowed to use the monotone convergence theorem but not the bounded or dominated convergence theorems)?
