# 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

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 $[a,b]$?

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 $[a,b]$.

• I'll look at the upper and lower sum version of Riemann Integration. We haven't used those but I should be able to translate it into the definition I gave. – oliverjones Jan 24 '17 at 3:55
• To answer your basic question -- yes, you need to prove that $f$ is integrable on $[a,b]$ before you manipulate limits with $r \to b$ as if this were an improper integral. You need a non-constructive approach as I have shown since you can't produce a specific candidate for the integral for general $f$ and show that Riemann sums converge to this value. You succeeded only in showing that integrals of $|f|$ over $[a,r]$ are bounded for $r < b$. – RRL Jan 24 '17 at 4:33
• I think the problem for me is we don't use infimums. Does it make sense if I say : choose $r$ such that $b-r < \frac{\epsilon}{4M}$ then $sup_{x \in [r,b]}\lvert f - f_\epsilon|(b-r) < 2M\frac{\epsilon}{4M} = \frac{\epsilon}{2}$ using my definition I gave above. Using the boundedness it makes sense to still use supremums with my definition. – oliverjones Jan 24 '17 at 4:35
• Adding to my last comment. It makes sense to me because the definition I gave seems like a shift. rather than having these two step functions that determine the upper and lower sums we shift down using the difference so the sup is determined by the $g_\epsilon$ and the inf is essentially the zero-constant step function so that $0 \le \lvert f - f_\epsilon \rvert \le g_\epsilon$ – oliverjones Jan 24 '17 at 5:09
• Your argument about $r$ is correct. Are you required to work with this alternative definition? – RRL Jan 24 '17 at 5:29

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.

• I haven't learned anything about measure theory or the Lebesgue Criterion. – oliverjones Jan 24 '17 at 3:00
• oh, you can check this out later then. – Jorge Fernández Hidalgo Jan 24 '17 at 3:01