Lebesgue integration on real line we know that if $A$ is a Lebesgue measurable set, $f$ and $g$ nonnegative, then 
$$ f \leq g \implies \int_A f dm \leq \int_A g dm $$
Does the result still follow if we change $\leq$ with $<$ ??

If $A$ is a null set, then we have of course
$$\int_A f\,dm = \int_A g\,dm = 0$$
regardless of $f$ and $g$, so then the strict inequality does not follow. But if $m(A) > 0$, then we have
$$f < g \Rightarrow \int_A f\,dm < \int_A g\,dm.$$
Let $A_n = \{x \in A : g(x) > f(x) + 2^{-n}\}$. Then $A = \bigcup A_n$, the $A_n$ are increasing, so $m(A_n) > 0$ for all large enough $n$. Then
$$\begin{align}
\int_A g\,dm &= \int_{A_n} g\,dm + \int_{A\setminus A_n} g\,dm\\
&\geqslant \int_{A_n} f + 2^{-n}\,dm + \int_{A\setminus A_n} f\,dm\\
&= \int_{A_n} f\,dm + 2^{-n} m(A_n) + \int_{A\setminus A_n} f\,dm\\
&= \int_A f\,dm + 2^{-n}\cdot m(A_n)\\
&> \int_A f\,dm.
\end{align}$$
