Let $(X,\mathcal{A},\mu)$ be a a measure space, and $f$ and $g$ two measurable functions. Now if $f$ and $g$ are nonnegative and $f\leq g$, it can be easily seen that $\int f\,d\mu\leq \int g\,d\mu$, where the possibility of either side being $\infty$ is allowed.
But if $f$ and $g$ are NOT nonnegative, is it still true? It is of course true if they are in $L^1$ but if we are not sure they are in $L^1$ is it true? What if we know nothing about $f$ (besides being measurable) but do know that $g$ is nonnegative or that $g\in L^1$, can we still say $\int f\,du\leq\int g\,d\mu$?
Any feedback will be welcome and appreciated, thanks in advance
We can say that $\int (g-f)\,d\mu\ge 0$. But to state the inequality with two integrals, we must first define what $\int f\,d\mu$ means when $f$ is not an $L^1$ function. Usually, one introduces $f^+=\max(f,0)$ and $f^-=\max(-f,0)$, so that both $f^+,f^-$ are nonnegative and $f=f^+-f^-$. Then if at least one of these two functions is in $L^1$, the right hand side of $$\int f\,d\mu = \int f^+\,d\mu-\int f^-\,d\mu$$ makes sense and serves as the definition of the left hand side. If both are infinite, we give up: no way to assign a meaning to $\infty-\infty$.
If both $\int f\,d\mu$ and $\int g\,d\mu$ are defined, the inequality holds. Indeed, $f\le g$ implies $f^+\le g^+$ and $f^-\ge g^-$. These are nonnegative, so we have
$$\int f^+\,d\mu \le \int g^+\,d\mu,\qquad \int f^-\,d\mu \ge \int g^-\,d\mu$$ It remains to subtract the inequalities: no $\infty-\infty$ appears.