2
$\begingroup$

i need to prove that if Riemann integrable in $[a,c]$ and $[c,b]$ then is RI in $[a,b]$, maybe is easy but i can't see it.

Definition of Riemann Integral
A function $f:[a,b]\rightarrow \mathbb{R}$ is Riemann Integrable on $[a,b]$ if
$\exists \thinspace L \in \mathbb{R} : \forall \epsilon >0\thinspace \exists \thinspace \delta >0 :$ if $P$ is any tagged partition of $[a,b]$ with $\|P\|<\delta$ then
$\|S(f,P)-L\|<\epsilon$ , where $S(f,P)=\sum_{i=1}^{n}f(t_{i})(x_{i}-x_{i-1})$ is Riemann sum, $t_{i}\in[x_{i-1},x_{i}]$ and $\|P\|:=\max\{|x_i-i_{i-1}|1\leq i\leq n\}$

Here is my trial:
Give $\epsilon>0$:
$f\in R([a,c])$ $\rightarrow \exists \delta'>0$ $\forall$ $P'=\{a=x_{0},...,x_{m}=c\}$ with $\|P'\|<\delta' \rightarrow$ $\big|\sum_{i=1}^{m}f(t'_{i})(x_{i}-x_{i-1})-L\big|<\epsilon/2$

$f\in R([c,b])$ $\rightarrow \exists \delta''>0$ $\forall$ $P''=\{c=x_{m},...,x_{n}=b\}$ with $\|P''\|<\delta ''\rightarrow$ $\big|\sum_{i=m+1}^{n}f(t''_{i})(x_{i}-x_{i-1})-K\big|<\epsilon/2$

Taking $\delta=min\{\delta',\delta''\}$ then for $\|P\|<\delta$, where $P$ is a partition $P=\{x_{1},...,x_{m},x_{m+1},...,x_{n}\}$ ; we have:$\big|\sum_{i=1}^{m}f(t'_{i})(x_{i}-x_{i-1})-L\big|<\epsilon/2$ $\thinspace$ and $\thinspace$ $\big|\sum_{i=m+1}^{n}f(t''_{i})(x_{i}-x_{i-1})-K\big|<\epsilon/2$

I'm thinking about using triangular inequality, but my question is what happens with the $t'_{i}$ and $t''_{i}$ how are these terms grouped or is it just a sum, please help. Thanks in advance

$\endgroup$
  • $\begingroup$ I see you are trying to prove this with Riemann sums rather than Darboux sums as in the duplicate so I will reopen and give you a hand. $\endgroup$ – RRL Apr 21 at 21:45
1
$\begingroup$

Consider any partition $P = (x_0,x_1,\ldots, x_n)$ of $[a,b]$ such that $\|P\| < \delta =\min (\delta', \delta'', \epsilon/(9M))$ where $M = \sup_{x \in [a,b]}f(x)$.

Suppose $c$ is not one of the partition points so that for some index $k$ we have $x_{k-1} < c < x_k$. For any choice of tags $\{t_j\}_{j=1}^n$ we have

$$\tag{*}\left|\sum_{j=1}^n f(t_j)(x_j-x_{j-1}) - (L + K)\right| \\ = \left|\sum_{j=1}^{k-1} f(t_j)(x_j-x_{j-1}) + \sum_{j=k+1}^{n} f(t_j)(x_j-x_{j-1})+f(t_k)(x_k- x_{k-1})- (L + K)\right| \\ \leqslant\left|\sum_{j=1}^{k-1} f(t_j)(x_j-x_{j-1}) + f(\xi')(c- x_{k-1})- L\right| +\left| f(\xi'')(x_{k}-c) + \sum_{j=k+1}^{n} f(t_j)(x_j-x_{j-1})-K\right| + \left|f(t_k)(x_{k}-x_{k-1}) - f(\xi'')(x_{k}-c) - f(\xi')(c-x_{k-1}) \right|$$

where $\xi'$ and $\xi''$ are arbitrary intermediate points for the intervals $[x_{k-1},c]$ and $[c, x_k]$.

Note that

$$\left|f(t_k)(x_{k}-x_{k-1}) - f(\xi'')(x_{k}-c) - f(\xi')(c-x_{k-1}) \right| \leqslant M(x_{k}-x_{k-1}) +M(x_{k}-c) + M(c-x_{k-1})<3M\delta < \epsilon/3$$

Each of the other two terms on the RHS of (*) is less than $\epsilon/3$ for an appropriate choice of $\delta'$ and $\delta''$ by your own argument.

If the point $c$ is a partition point in $P$, then the argument is similar but even simpler.

$\endgroup$
  • $\begingroup$ Ok i understand the last equation is limited by $M$ and $\delta$ and taking $\delta$ proper as $\delta<\epsilon/9M$ then $\delta\leq\{\delta',\delta'',\epsilon/9M\}$. TY so much!!! $\endgroup$ – user665960 Apr 22 at 2:06
  • $\begingroup$ @user665960: You're welcome. $\endgroup$ – RRL Apr 22 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.