Prove for bounded $f$, if $f:[a,r] \to \mathbb{R}$ is Riemann-Integrable for $r \in [a,b)$ then $f:[a.b] \to \mathbb{R}$ is Riemann-Integrable

Solution 1:

For any $ r \in [a,b)$, $f$ is Riemann integrable, and for any $\epsilon > 0$ there exists a partition $P_r$ of $[a,r]$ such that by the Riemann criterion, the difference between upper and lower sums satisfies

$$U(P_r,f) - L(P_r,f) < \epsilon/2.$$

Since $f$ is bounded, we can choose $r$ such that $b-r < \epsilon/(4M)$ and

$$[\sup_{x \in [r,b]}f(x) - \inf _{x \in [r,b]}f(x)](b-r) < 2M\frac{\epsilon}{4M} = \frac{\epsilon}{2} ,$$

where $M = \sup_{x \in [a,b]}|f(x)|$.

Extending the partition $P_r$ to a partition $P$ of $[a,b]$ by adding the point $b$ we have

$$U(P,f) - L(P,f) = U(P_r,f) - L(P_r,f) + [\sup_{x \in [r,b]}f(x) - \inf _{x \in [r,b]}f(x)](b-r)< \epsilon.$$

Thus $f$ is Riemann integrable on $[a,b]$.

Solution 2:

proof with measure theory:

Let $X$ be the set of discontinuities of $f$.

Let $X_n=X\cap [a,b-\frac{1}{n}]\cap X$. Notice that $X\subseteq \{1\}\cup\bigcup\limits_{i=1}^\infty X_n$. By hypothesis this is a countable union of sets with lebesgue measure zero. We conclude $X$ has measure zero, and so $f$ is integrable on $[a,b]$ by Lebesgue's criterion.