Let $A$ be a bounded, path-connected subset of $\mathbb{R}^n$ where $\partial A$ has measure zero. Let $f:A\to\mathbb{R}$ be a bounded, continuous function. Show that there exists $a\in A$ such that $I:=\int_A f\, dV=f(a)\text{vol}(A)$.
My attempt:
Let $R$ be a rectangle containing $A$. Let $\bar{f}=f$ on $A$ and $\bar{f}=0$ otherwise. Let $m=\inf f(A)$ and $M=\sup f(A)$. We want to show that $m\le I/\text{vol}(A)\le M$. We first attempt to show that $I\le M\text{vol}(A)$. It suffices to show that for any partition $P$ of $R$, $$ U(\bar{f},P)=\sum_{Q\in P}\sup \bar{f}(Q) \text{vol}(Q) \le M \sum_{Q\in P}\inf 1_A(Q)\text{vol}(Q)=M\cdot L(1_A,P).$$ For each $Q\in P$, if $Q\subset A$, then $\sup\bar{f}(Q)=\sup f(Q)\le\sup f(A)\le M\inf 1_A(Q)$. However, when $Q$ crosses the boundary, we no longer have this easy inequality. I am struggling on how to apply either that $\partial A$ has measure zero or that $A$ is path-connected here.
Also, after we have shown that $m\le I/\text{vol}(A)\le M$, if it happens that $I/\text{vol}(A)=M$ or $m$, how should one apply the path-connectedness here? We do not have compact domain here.