Prove a simple property of Lebesgue integrals

Question

Given $B \subset A$, can we prove that $\int_{B} fdp < \int_{A}fdp$? If yes how? What assumptions should be made in order for the inequality to be true? Intuitively it makes sense to me, because it's like integrating a smaller area (loosely speaking) but I could not find any known property covering this case. Thank you.

Answer 1

I suppose that for you $B\subset A$ means $B\subsetneq A$.

As a trivial counterexample you could consider the zero function. In general the inequality doesn't have to be true because you could have $\displaystyle\int_{A\setminus B}f\,\text{d}p=0$.

As Teresa Lisbon rightly pointed out in the comments, you would have the inequality if $f>0$ on $A\setminus B$ and $A\setminus B$ is a set of positive measure, i.e., $p(A\setminus B)>0$, since then $\displaystyle\int_{A\setminus B}f\,\text{d}p>0$ (the Lebesgue integral of a positive function on a set of positive measure is positive). Therefore $\displaystyle\int_A f\,\text{d}p=\int_{A\setminus B}f\,\text{d}p+\int_Bf\,\text{d}p>\int_Bf\,\text{d}p$