If $(E[\int_0^T|f_n(s)|^2ds])_n$ converges, does it exists a subsequence $f_{n(j)}$ such that the integral converges for $t<T$?

Question

Suppose $f, f_1, f_2, \dots: [0, T] \times \Omega \to \mathbb R$, and that $$ \lim_n \mathrm E\left [ \int_0^T \left |f_n(s, \cdot) \right|^2 ds\right ] \to \mathrm E\left [ \int_0^T \left |f(s, \cdot) \right|^2 ds\right ] < \infty. $$

Does it follow that for any $0 < t < T$, there exists a subsequence $n(j)$ such that $$ \lim_j \mathrm E\left [ \int_0^t \left |f_{n(j)}(s, \cdot) \right|^2 ds\right ] \to \mathrm E\left [ \int_0^t \left |f(s, \cdot) \right|^2 ds\right ]\quad ? $$ If so, how would you prove it?

Must grateful for any help provided!

Answer 1

No, it does not!

$T=1$, $f_n = 1$ everywhere and $f(t,\cdot)=t\sqrt{3}$, then $$ \mathbb{E}[\int_0^1|f_n(s,\cdot)|^2ds] = 1 \stackrel{n \to \infty}\to 1 = \mathbb{E}[\int_0^1|f(s,\cdot)|^2ds]$$

But $$ \mathbb{E}[\int_0^t|f_n(s,\cdot)|^2ds] = t \stackrel{n \to \infty}{\not\to} t^3 = \mathbb{E}[\int_0^t|f(s,\cdot)|^2ds]$$