$$ \textbf{Problem} $$ $\int_A f dm \leq 0 $ for all $A$ lebesgue measurable set implies $f \leq 0 $ a.e
$$ \textbf{Solution (Attempt)} $$
We want to show $X = \{ x : f > 0 \} $ is a null set, that is $m(X) = 0 $. Consider, $X_n = \{ x : f \geq \frac{1}{n} \} $. We show
$$ X = \bigcup_{n=1}^{\infty} X_n $$
If $x \in X$, then $f > 0 $. By archimidean, can find $n$ such that $f \geq \frac{1}{n}$. therefore, $x \in X_n $ for some $n$. $\therefore$ by definition $x \in \bigcup_n X_n $.
For the reverse direction, if $x \in \bigcup_n X_n $, then $x \in X_n$ for some $n$. Hence, $f \geq \frac{1}{n} > 0 \implies f > 0 \implies x \in X $
Now, we show $m(X_n) = 0$. but by hypothesis,
$$ \int\limits_{X_n} \frac{1}{n} \leq \int\limits_{X_n} f \leq 0 \implies \frac{1}{n} m(X_n) \leq \int\limits_{X_n} f \leq 0 $$
This can only happen if $m(X_n) = 0$ Which implies by additivity of the measure that $m(X) = 0$. Therefore, $f \leq 0 $ a.e.
Is this correct?
Put $E = [f > 0]$. This is measurable. Since $f >0$ on $E$, $$\int_Ef(x)\,dx \ge 0.$$ but by your hypothesis, this integral cannot be positive, so $$\int_E f(x)\,dx = 0$$ Necessarily, $f = 0$ a.e. on $E$, so $|E| = 0$. Since $[f>0]$ has measure zero, $f \le 0$ a.e.