Let $T>0$. Let $f:[0, T]\times \mathbb R \rightarrow [0, \infty[$ such that $f\in L^1([0, T]\times \mathbb R)$. I want to prove that there exists a set of full measure $J\subseteq [0, T]$ such that for all $t\in J$, $$\int_\mathbb R f(t, x)dx \geq C,$$ for a fixed positive constant $C$.
I have proved that for all $g\in C^\infty([0, T]\times \mathbb R)$ and for all $t\in [0, T]$, $$\int_\mathbb R g(t, x)dx \geq C.$$
Is there any way of approximating $f$ by functions of $C^\infty([0, T]\times \mathbb R)$ in order to prove the statement?
As already pointed out $f \equiv 0$ is a counter-example. Even if you avoid this trivial case by assuming that $\int_0^{T}\int_{\mathbb R} f(t,x) dxdt >0$ it is still false. Take $f(t,x)=t^{2}I_{(0,\frac 1 t)}(x)$ for a counter-example.