Let $(X, \mu)$ be an infinite measure space. Let $f_n:X \to \mathbb{R}$ be a sequence of uniformly bounded, continuous, non-negative functions on $X.$ Let $f_n \to 0$ uniformly.
Does this necessarily imply:
$$\int_{X}f_nd\mu \to 0?$$
If not, what other criteria do we need to ensure this? For the moment, some hints in the comments would be enough I think, as I'd like to work it out.
On
As written in the opening post, the convergence $\int_{X}f_nd\mu \to 0$ may not hold. PhoemueX gave the example $f_n \equiv 1/n$.
One has to add the assumption that for each positive $\varepsilon$, there exists a set $A=A(\varepsilon)$ of finite measure such that $$ \sup_{n\geqslant 1}\int_{X\setminus A} f_n d\mu<\varepsilon. $$ Then for a positive $\varepsilon$ and the corresponding $A$, $$ \int f_nd\mu\leqslant \varepsilon +\int_Af_nd\mu\leqslant\varepsilon+\mu(A)\sup_{x\in X}f_n(x). $$
No, it doesn't imply.
Consider $X = \mathbb{R}$ and the sequence of functions such that
$f_i(x) = 1/(i+1) $ for $-(i+1)<x<(i+1)$ and $0$ otherwise. You can easily make each $f_i$ continuous.
Everything is well-defined here: All $f_i$ are integrable and the result of integral is finite for all $i$; the function sequence converges and so does the integral sequence.
However, the function sequence converges to zero uniformly while the integrals don't.