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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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