Problem
On Probability space ($\Omega$, $\mathscr{A}$,P), I have sequence of $L_1$ non-negative function $f_i$ s.t.$\int_A f_i dP$ converge and $\int_A f_i dP$ is bounded by $1$ for every A $\in$ $\mathscr{A}$. The problem is the only thing I know is convergence and I do not know whether or not it converge to $\int_A f dP$ for some $f$. Is there any chance I can conclude $\int_A f_i dP$ converge to $\int_A f dP$ for some $f$?
I have tried
construct such $f$ by proving some subsequence of $f_i$ is pointwise convergence to some $f$ almost everywhere.
prove $f_i$ is Cauchy in $L_1$, so that this also imply existence of such $f$ by completeness and convergecne in $L_1$ gives us (1).
construct $f$ using simple function, but this seems difficult in abstract space $\Omega$.
None of these successed. This might be a stupid question. Could anyone help me on this? Thanks！
For each $A \in \mathscr{A}$, since $\int_A f_i dP$ converge, let $\mu(A)= \lim_{i \to \infty} \int_A f_i dP$.
We can prove that $\mu$ is a measure (and since $\int_A f_i dP$ is bounded by $1$ for every A $\in$ $\mathscr{A}$, $\mu$ is in fact a probability). One way to prove that $\mu$ is countably additive is to apply Vitali–Hahn–Saks theorem.
Now, it is easy to see that $ \mu \ll P$, so by Radon–Nikodym theorem there is a finite measurable function $f$, such that, for all $A \in \mathscr{A}$,
$$ \int_A f dP = \mu(A) =\lim_{i \to \infty} \int_A f_i dP$$