Integral inequality of measurable functions for every measurable set implies function inequality a.e. as long as measure is $\sigma$-finite

Question

Let $(X, \mathscr{A}, \mu)$ be a measure space and let $f,g: E \to [0, \infty]$ be measurable functions such that: $\int_A f d\mu \leq \int_A g d\mu$ for every measurable $A \subseteq E$. If $E$ has $\sigma$-finite $\mu$-measure, then $f \leq g$ $\mu-$a.e. in $E$.

I am not sure how to go about showing this. I have seen, in the case of the Lebesgue measure, that two functions being equal almost everywhere implies that their integrals are equal. However, this proposition is going the other direction and is extended to general measure spaces.

I am also confused about the need for a $\sigma$-finite $\mu$-measure. Presumably, the need for this comes up somewhere in the proof of the statement, but as I am not sure how to begin, it is not clear to me why this is necessary.

I would appreciate any hints to get the ball rolling on this question or references to textbooks covering this.

Answer 1

Since $\mu$ is $\sigma$-finite, we can can write $X = \bigcup_{n=1}^\infty X_n$ with $\mu(X_n) < \infty$. We only need to show that $f(x) \leq g(x)$ for $\mu$-almost all $x \in X_n$ and all $n \in \mathbb{N}$.

Thus, let w.l.o.g. $\mu$ be a finite measure. Now let $Y_n := \{ g \le n\}$. Again, we only need to show the that $f(x) \le g(x)$ for almost all $x \in Y_n$ and $n \in \mathbb{N}$.

Therefore we can also assume additionally that $g$ is bounded, i.e. integrable. Now let $A_\varepsilon := \{f \ge g + \varepsilon\}$. The condition implies $$\int_{A_\varepsilon} g \, d \mu \ge \int_{A_\varepsilon} f \, d \mu \ge \varepsilon \mu(A_\varepsilon)+\int_{A_\varepsilon} g \, d \mu. $$ Since both sides are finite, this can only happen if $\mu(A_\varepsilon) =0$. All in all, we find $$ \mu(f > g) = \mu( \bigcup_{n=1}^\infty \{f \ge g +1/n\}) \le \sum_{n=1}^\infty \mu(A_{1/n}) =0. $$

Answer 2

It is easy to reduce the proof to the case when $E$ has finite measure. Let $f_N=\min \{f,N\}$ and $g_N=\min \{g,N\}$. If we show that $f_N \leq g_N$ a.e. for each $N$ we are done. Replacing $A$ in the hypothesis by $A_N\equiv A\cap \{f\leq N, g \leq N\}$ we get $\int_{A_N} f_N \leq \int_{A_N} g_N$. In particular $\int_{A_N} (g_N-f_N) \geq 0$ when $A=\{g<f-\epsilon\}$ [Note that $\infty - \infty$ does not appear here since the integrals are finite]. Can you now show that $A_N$ has measure $0$ (for every $\epsilon >0$) from this an conclude that $f_N \leq g_N$ a.e.?