# Function is Riemann integrable in $[a,c]$ and $[c,b]$ then is RI in $[a,b]$

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

• 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. – 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.

• 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!!! – user665960 Apr 22 at 2:06
• @user665960: You're welcome. – RRL Apr 22 at 2:24