Comparison between the Lebesgue integral of a given function over two different sets.

Question

Amid one of my studies, I came across a slight problem. I believe this should be a trivial question for someone that has a strong background in Lebesgue integral theory.

Here it goes:

Let's say we have a certain Lebesgue-integrable function $f$ and two sets $A$ and $B$ such that

$$ f|_A \leqslant f|_B $$

Is it true that

$$ \int_A f(x) dx \leqslant \int_B f(x) dx ? $$

Note. I am not sure if the notation is clear, since this question was made up by myself. When I state $f|_A \leqslant f|_B$ I mean that for every $x \in A$ and for every $y \in B$ we have that $f(x) \leqslant f(y).$

Answer 1

No, it is not true. You would additionally need that $m(A) \leq m(B)$. For a counterexample, consider $f \equiv 1$ and $A=[0,1]$, $B=\{2\}$.

Edit: with the additional assumption, we can prove the claim as follows. Let $L = \sup_A f$ and $M = \inf_B f$; your "$f|_A \leq f|_B$" hypothesis states that $L \leq M$. Then $$ \int_A f \leq L \cdot m(A) \leq M \cdot m(B) \leq \int_B f. $$