limsup in an integral inequality

Question

Can someone explain why

$ \int_D f \geq limsup_{n \rightarrow \infty} \int_D f_k$

holds if $f_k \rightarrow f$ almost every where thus to say the set of x's where it isn't the case is a Nullset?

Answer 1

This is false. Let $f_k = k I_{(0,\frac 1 k)}, f\equiv 0$. Then $f_k(x) \to f(x)$ for every $x$ but the inequality says $0 \geq 1$.