Is $f$ integrable on the Jordan domain $V$? If $D$ is a bounded subset of $\mathbb{R^n}$ and $f$ is integrable on $D$, $V$ is a Jordan domain containing in $D$, is $f$ integrable on $V$?
I think it is true. I tried to consider the restriction function of $f$ on V, but I don't know how to discuss. And I think maybe we can use the contradiction.
 A: Given the usual definition of the Riemann integral, to show that $\displaystyle\int_Vf$ exists, you need to show that $\displaystyle\int_Rf\cdot\chi_V$ exists for some closed rectangle $R \supset V$. (The choice of $R$ does not matter.)

Let us fix some closed rectangle $R \supset D$. (Note that $D$ is bounded.)
Since $\displaystyle\int_Df$ exists, we see that the set of discontinuities of $f\cdot\chi_D$ is of measure zero.
Now, we show that the same is true for $f\cdot\chi_V$.
Let us consider the three different sets where $x \in R$ can be:

*

*$x \in \operatorname{int}V$.
In this case, $x \in \operatorname{int}D$ as well thus, and in a neighbourhood of $x$, we have that
$$f = f\cdot\chi_V = f\cdot\chi_D.$$
Thus, if $f\cdot\chi_V$ is discontinuous at $x$, then so is $f\cdot\chi_D$.
Since the set of discontinuities of the latter has measure zero, we see that
$$D_1 := \{x \in \operatorname{int}V \mid f\cdot\chi_V \text{ is discontinuous at }x\}$$
also has measure zero.

*If $x \in \operatorname{boundary}V$.
Since $V$ is a Jordan domain, the set of all such $x$ itself has measure zero. In particular, so does the set
$$D_2 := \{x \in \operatorname{boundary}V \mid f\cdot\chi_V \text{ is discontinuous at }x\}.$$

*$x \in R \cap \operatorname{int}(\Bbb R^n\setminus V).$
$f\cdot x_V$ is continuous at any such $x$ as it is identically $0$ in a neighbourhood of $x$.


Thus, we see that the complete set of discontinuities of $f$ is $D_1 \cup D_2$ which has measure zero. Thus, $f$ is integrable on $V$.
