When does convergence in distribution imply convergence of integrals?

Question

Suppose $f_n \rightarrow f$ in distribution, where $(f_n)_{n \geq 1}$ is a sequence of integrable functions. Also suppose $\sup_{n \geq 1} |f_n|$ is well-defined and measurable. Does this imply that

$$ \int_0^1 f_n \rightarrow \int_0^1 f $$

I've tried thinking about convergence of the characteristic functions (Levy's continuity theorem) and also supposing that $f_n \rightarrow 0$ in distribution. But I couldn't get to the result.

I don't think it is true given my attempts, since potentially you could use $f_n = n \chi_{[0,1/n]}$, which are (i) integrable and (ii) have well-defined/measurable supremum on the interval (0,1). But $\int_0^1 f_n = 1$ and $f_n \rightarrow \chi_{(0)}$ in distribution.

Answer 1

On the open interval $(0,1)$ with Lebegue measure define $f_n(x)=n1_{(0,\frac 1n)}(x)$ and $f=0$ Then $f_n(x)\leq \frac 1 x$ for all $n,x$ but $\int f_n$ does not tend to $\int f$. Note that $f_n(x) \to f(x)$ for all $x$.