If $m((A \setminus B) \cup (B\setminus A))=0$ then $\int_{A} f(x) dx= \int_{B} f(x) dx$ for every non negative measurable function $f$.
How to show this is true? I thought of arguing as follows. $A= (A\setminus B) \cup (A \cap B)$ and $B= (B\setminus A) \cup (B \cap A)$
Now $m(A \setminus B)=0$ which implies $m(A)= m(A \cap B)= m(B)$ This means for indicator functions, the two integrals coincide. Then by linearity the result holds for simple functions. Then by a limiting argument since every non negative measurable function can be approximated by an increasing sequence of simple functions, the result holds by monotone convergence theorem. However, writing it out I was not able to rigorously prove it. Can somebody help me out? Thanks.
Let it be that $f$ is a measurable nonnegative function such that $m(\{x\mid f(x)>0\})=0$.
Claim: $\int f=0$.
Proof: If $\int f>0$ then (on base of the definition of the integral) a measurable function $g=c_11_{A_1}+\cdots+c_n1_{A_n}$ must exist with $0\leq g\leq f$ and $c_1m(A_1)+\cdots+c_nm(A_n)>0$. However, we have $A_i\subseteq\{x\mid f(x)>0\}$ and consequently $m(A_i)=0$ for every $i$, so a contradiction is found.
Corollary: if $m(C)=0$ then $\int_Cf=0$. This because $\int_Cf=\int f1_C$ by definition and $\{x\mid f1_C(x)>0\}\subseteq C$ so that also $m(\{x\mid f1_C(x)>0\})=0$.
Applying this you can find in your situation (where $m(A\setminus B)=m(B\setminus A)=0$):$$\int_Af=\int_{A\cap B}f+\int_{A\setminus B}f=\int_{A\cap B}f=\int_{A\cap B}f+\int_{B\setminus A}f=\int_Bf$$