Does there exist a bounding integrable continuous function?

Question

Assume we are on the space $[a,b]$ and we have a function $f\in L^1([a,b])$. Then there exists a sequence of continuous functions $\{f_n\}$ converging to $f$ in $ L^1([a,b])$. Does it also exist a continuous function $g\in L^1([a,b])$ such that $|f_n(x)|\le g(x)$?

Answer 1

There is a continuous function $f_n$ such that $f_n(x)=\frac n {\ln n }$ on $(0,\frac 1 n)$, $0$ on $(\frac 2 n , 1)$ and linear in $[\frac1 n , \frac 2 n]$. In this case $f_n \to 0$ on $L^{1}(0,1)$ but there is no integrable function $g$ such that $f_n \leq g$ a.e. for every $n$. Proof: On $(\frac 1{n+1}, \frac 1 n)$ we have $g(x) \geq \frac n {\ln n }$ so we get $\int_0^{1} g(x) dx \geq \sum_n \frac n {\ln n } (\frac 1n -\frac 1 {n+1})=\infty$.