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?
Let $f,g$ be both non-negative measurable functions and define: $$ E_n=\{x: g(x)\geq f(x)+t_n\}, $$ where $t_n$ is a sequence of decreasing positive numbers converging to $0$ as $n\to\infty$. For this choice, we have: $$ \mu(E_n)=\int_{E_n}d\mu\leq \int_{E_n}\frac{g(x)-f(x)}{t_n}d\mu=\int_{E_n}\frac{g(x)}{t_n}d\mu-\int_{E_n}\frac{f(x)}{t_n}d\mu\leq 0, $$ where we used the linearity of integration. To see why the linearity of integration holds note that $g(x)-f(x)$ is non-negative over $E_n$ and so are $g$ and $f$. You can define $h_n=(g-f)1_{g-f\geq t_n}$ and do the steps with $h_n$.
Therefore $\mu(E_n)$ is zero and so is $\mu(\cup E_n)=0$, implying that: $$ \mu(\{x: g(x)> f(x)\})=0. $$