Let f be a real-valued function on a subset $A$ on $E^n$ and let $B \subset A$. Show that if $\int_A f$ exists and $B$ has volume, then $\int_B f$ exists.
Note: Volume is the Jordan measure. Integration is Riemann type.
Edit: I deleted my work for the first part as I assumed it was correct. I just need help with the above.
Will this work?
$$\int_A f = \int_{E^n} f \Bbb 1_A $$ $\Bbb 1_B$ exists because $B$ has volume ($\Bbb 1$ is indicator function). Thus, $$\int_{E^n} f \Bbb 1_B $$ exists. And, $$\int_{E^n} f \Bbb 1_B = \int_B f $$
This seems too easy.
Sketch of proof:
Let assume that the set A is bounded, so it can enclosed in a rectangle. Now, we know that function $g:A \to \mathbb{R}$ is Riemann-integrable iff set of discontinuities of $g$ forms a set of measure zero (i.e. for all $\epsilon > 0$ we can find a disjoint finite rectangle cover whose jordan measure is less than $\epsilon$).
We know that $f:A \to \mathbb{R}$ is Riemann-integrable, so its' set of discontinuities has zero measure. Furthermore, set $B \subset A$ has a volume. Which means that $1_B$ is Riemann-integrable and thus its' set of discontinuities has also zero measure.
To show that $\int_B f$ exists it is enough to show that discontinuities of function $1_B \cdot f$ has zero measure.
The main idea is to enclose discontinuities of both function separately with families of disjoint rectangles, $C_1, C_2$ with jordan measures of $m(C_1),m(C_2) \leq \frac{\epsilon}{2}$. Then forming a new family $C_3 = C_1 \cup C_2$. Functions $f$ and $1_B$ are continuous outside of set $C_3$, thus, their product is also continuous. Now, Measure of the set $C_3$ is at most $m(C_1) + m(C_2) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$. Thus, discontinuities of $1_B \cdot f$ form a set of zero measure.
This is just a sketch of proof. To be more precise we need to be more careful about: