Prove that a Jordan measurable set of measure 0 is Riemann integrable

Question

My problem is: I have a set $A$ which is Jordan measurable, and I have a function $f:A\rightarrow\mathbb{R}$. I need to prove that if the measure of A is $0$, then $f$ is Riemann integrable and furthermore $\int_{A}f(x)dx = 0$.

Thank you!

Answer 1

Since $A$ is Jordan measureable it is bounded. The Riemann integral over $A$ is defined as $\displaystyle \int_Q f \chi_A$ where $Q$ is a rectangle containing $A$.

(1) Prove first that the integral of $f\chi_A$ over $Q$ actually exists by showing that the integrand is continuous except on a set of measure $0$.

(2) Take any partition $P$ of $Q$. Every subrectangle $R$ of $P$ has nonzero content and cannot be contained in $A$. Thus, $R$ always contains a point where $f(x)\chi_A(x) = 0$ and

$$\inf_{x \in R} f(x)\chi_A(x) \leqslant 0 \leqslant \sup_{x \in R} f(x)\chi_A(x), $$

This implies that lower and upper Darboux sums satisfy $L(P, f\chi_A) \leqslant 0 \leqslant U(P,f \chi_A )$. From here you should be able to prove that the integral (which falls between lower and upper sums) is $0$.