Let $(\Omega,\mathcal{F},\mu)$ be a measure space and $f,g:\Omega\to[0,\infty]$ be two non-negative measurable functions. If $\forall A\in\mathcal{F}: \int_A fd\mu\geq \int_A gd\mu$, can we conclude $f\geq g \, \mu-$a.e.?
I know that in case $f,g:\Omega\to\mathbb{R}$ and $f,g\in L^1(\mu)$ the result holds, since $\infty-\infty$ does not happen and the theorem becomes equivalent to $\forall A\in\mathcal{F}: \int_A f-gd\mu\geq0\implies f-g\geq0\, \mu-$a.e., which I know the proof.
Is this still true without $L^1$ assumption for non-negative measurable functions, where integral can become $+\infty$ and how can I prove that?