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

Question

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 and $\displaystyle \lim_{r \to b, r < b} \int_a^r f= \int_a^bf$.

Alternative definition for Riemann-Integrable functions :: $f : [a,b] \to \mathbb{R}$ is Riemann-Integrable if $\forall \varepsilon > 0, \exists$ step functions $f_\varepsilon, g_\varepsilon$ such that $\lvert f- f_\varepsilon \rvert \le g_\varepsilon$ and $Ig_\varepsilon \le \varepsilon$ where $Ig_\varepsilon = \sum_{j = 0}^{N}g_\varepsilon(x)(\Delta x_{j+1} - \Delta x_j)$.

$f$ is bounded so $\displaystyle \exists M \in \mathbb{R}^+$ s.t. $|f(x)| \le M, \forall x \in [a,b] $ so we know $\displaystyle|\int_a^r f| \le \int_a^r|f| \le M(r-a) \le M(b-a) $. Now, for some index set $I$ where $r_0 \le r_1 \le \ldots\le r $ we have $\displaystyle \int_a^rf = \sum_{i \in I}\int_{a}^{r_{i}}f$

Do I need to show $f$ is Riemann-Integrable over $[a,b]$ before I can solve the limit? It seems like it but seems like the other way would give it to me.

EDIT:

Since $f(x) \le M, \forall x \in [a,b]$ and we have there exists $r \in [a,b]$ such that $b - \frac{1}{n+1} < r < b$ we have $\displaystyle \int_a^{b-\frac{1}{n+1}}|f| \le \int_a^r|f| \le M(b-a)$ Does this give me that $f$ is Riemann-Integrable over $[a,b]$?

Answer 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 $[a,b]$.

Answer 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.