If $f:A\rightarrow \mathbb{R}$ is an integrable function on $B$, do $\int_Af$ exists? Let $A$ and $B$ be open sets of $\mathbb{R}^n$ such that $A-B$ has measure zero. If $f:A\rightarrow \mathbb{R}$ is an integrable function on $B$, can we ensure that $\int_Af$ exists?
I am trying show that $\int_Af$ exists, where
$$A=\{(x, y)\in \mathbb{R}^2: x>0\text{ and } y>0\}$$
and the function $f:A\rightarrow \mathbb{R}$ is given by
$$f(x)=\frac{1}{(x^2+\sqrt{x})(y^2+\sqrt{y})}.$$
I proved that $\int_Bf$ exists, where
$$B=\{(x, y)\in A: x\neq 1\text{ and } y\neq 1\}.$$
We have that $A-B$ has measure zero, so we can ensure that $\int_Af$ exists?

You are right. You can use that the integral over $A$ is equal to the integral over $A\cap B$ plus the integral of $A\setminus B$ that is zero (and that $f\geq 0$). So $f$ is an integral function over A.
