A Measureable Function is less than or equal to another iff all its integrals are less than or equal to the others
# Statement
Let $(X, \mathcal{M}, \mu)$ be a Measure Space and let $f, g \in L^{1}(\mathcal{M})$. Then $f \leq g$ Almost Everywhere If and Only If $$\int\limits_{A} f d \mu \leq \int\limits_{A} g d \mu$$ $\forall A \in \mathcal{M}$.
# Proof
$(\Rightarrow)$: Let $A \in \mathcal{M}$. If $f \leq g$, then $\mathbb{1}{A}f \leq \mathbb{1}{A}g$. By Integration is Non-Decreasing, we get the result. $\checkmark$
$(\Leftarrow)$: Consider $A = [f > g] \in \mathcal{M}$. Observe that $\int\limits_{A} f - g =\int\limits_{A} f - \int\limits_{A} g \leq 0$. But $\forall x \in A$, $f(x) - g(x) > 0$, so $\int\limits_{A} f - g \geq 0$. Therefore $\int\limits_{A} f - g = 0$. Recall that Integration of a Nonnegative Function is 0 iff the Function is 0 Almost Everywhere, so we must have that $\mathbb{1}{A} (f - g) = 0$ Almost Everywhere. But $\forall x \in A$, $\mathbb{1}{A} (f(x) - g(x)() > 0$, therefore $\mu(A) = 0$. Thus $f \leq g$ Almost Everywhere. $\blacksquare$