Let $(X,\mathfrak{U},\mu)$ be a measurable space and $f,g\colon X\to\overline{\mathbb{R}}$ be integrable with $$ \int\limits_{A}f\, d\mu\leq\int\limits_{A}g\, d\mu~\forall A\in\mathfrak{U}.~~~~~~~~~~(*) $$ Show that then $f\leq g~~~~~\mu-\mbox{a.e.}$
My proof is the following.
As integrable functions, $f$ and $g$ are measurable, i.e. $\mathfrak{U}\setminus\mathcal{B}(\mathbb{\overline{R}})$ - measurable. Then $f-g$ is measurable, too, because the sum resp. difference of measurable functions is itself measurable. So $$ (f-g)^{-1}((0,\infty])=\left\{x\in X: f(x)>g(x)\right\}=\left\{f>g\right\}=:M\in\mathfrak{U} $$ and $$ (f-g)^{-1}(1/n,\infty])=\left\{f>g+\frac{1}{n}\right\}=:M_n\in\mathfrak{U}. $$
So I can integrate over $M_n$, getting $$ \int\limits_{M_n}f\, d\mu\geq\int\limits_{M_n}\left(g+\frac{1}{n}\right)\, d\mu=\int\limits_{M_n}g\, d\mu+\int\limits_{M_n}\frac{1}{n}\, d\mu=\int\limits_{M_n}g\, d\mu+\frac{1}{n}\mu(M_n). $$ Because of (*), it is $$ \int\limits_{M_n}g\, d\mu+\frac{1}{n}\mu(M_n)\geq\int\limits_{M_n}f\, d\mu+\frac{1}{n}\mu(M_n). $$ So it is $$ \int\limits_{M_n}f\, d\mu\geq\int\limits_{M_n}f\, d\mu+\frac{1}{n}\mu(M_n). $$ Am I right, that this only can be right, if $\mu(M_n)=0$? If yes: why? The only reason I can find is, that the integral from $f$ over $M_n$ is $<\infty$, because $f$ is integrable and so the integral is in $\mathbb{R}$ and for real numbers, this can only be right, if $\mu(M_n)=0$.
Maybe you can give me a better reason?
Anyhow, if $\mu(M_n)=0$, then the continuity of $\mu$ from down says $\mu(M)=0$, because $M_n\uparrow M$.
I think this is pretty much the same as what you did, just a little shorter write up.
It is enough to show that if $\int_A f \ge 0$ for all measurable $A$, then $f \ge 0$ almost everywhere. Assume the contrary: there exists a measurable set $A$ of positive measure such that $f < 0$ on $A$. Then $$\int_A f \, d\mu < 0,$$ which is a contradiction.
To prove that $\int_A f \, d\mu < 0$, you can write $A$ as $A = \bigcup_{k=1}^n \{x \in A : f(x) < -1/k\}$. Then one of the sets $A_k = \{x \in A : f(x) < -1/k\}$ must have a positive measure (otherwise $A$ would have measure $0$), and clearly $\int_A f \, d\mu \le \int_{A_k} f \, d\mu < 0$.