Let $f$ be a non-negative measurable function. Prove that for $A$ and $B$ measurable subsets of $\mathbb{R}$, the following holds: $$ \int_{A \cup B} f \, dm + \int_{A \cap B} f \, dm =\int_{B} f \, dm+ \int_{A } f \, dm $$ my attempt As $A$ and $B$ are not disjoint, my idea was to use properties of measurable functions, that is, if $A$ is a subset of $B$, then the $\int_A f \, dm$ $\leq$ $\int_B f \, dm$. In this case, $A$ is a subset of $A \cup B$ and $B$ is a subset of $A \cup B$, so the $\int_A f \, dm$ $\leq$ $\int_{A \cup B} f \, dm$, and $\int_B f \, dm$ $\leq$ $\int_{A \cup B} f \, dm$ so
$$\int_A f \, dm + \int_B f \, dm \leq 2\int_{A \cup B} f \, dm$$
and $A\cap B$ is a subset of $A$, $A\cap B$ is a subset of $B$ so
$$2\int_{A \cap B} f \, dm \leq \int_A f \, dm + \int_B f \, dm$$
I don't see a way to handle this, and I'm not sure if I'm doing it right. Any help is welcome