Assume that $(f_n)$ is a sequence of Lebesgue integrable functions on $(0,1)$ such that \begin{align*} (1) &\lim_{n\rightarrow\infty}\int_0^1f_n \quad \textrm{exists}, \\ (2) &\lim_{n\rightarrow\infty} f_n(x) = f(x), \quad \textrm{for every}\quad x\in(0,1), \end{align*} for some function $f$.
Do the above conditions imply that $f$ is integrable and $\lim_{n\rightarrow\infty}\int_0^1 f_n = \int_0^1 f$ ?
I know that only (2) is not enough, but I think that combining it with (1) should suffice. I can't prove it, though. Any hints are appreciated!
Edit: The answer to my question is negative, the counterexample being $f_n=n\mathbb{1}_{(0,1/n)}$ and f=0, as pointed out by geetha290krm in the comment.