Proof that Integral a to b = Inegral a to c + Integral c to b I have to proof using the definition of the limit of the Riemann sum that:
($c$ is somewhere between $a$ and $b$)
$$\int_{a}^{b} f(x)\,\mathrm{d}x = \int_{a}^{c} f(x)\, \mathrm{d}x  + \int_{c}^{b} f(x)\, \mathrm{d}x$$
I want to take the limit of Riemann for the left expression, since we know that $C$ is in somewhere in the middle of $A$ and $B$, we can divide the limit of sum such that the first lim SUM goes from $a$ to c + lim SUM goes from $c$ to $b$, and then from these 2 limits and I can write that Integral from $a$ to c + Integral from c to b = Integral from $a$ to $b$. I can also use the figure to show what this really means.
Does this considers a proof? 
 A: If you consider that f is R-integrable on [a, b] and if $a$ $<$ $c$ $<$ $b$,then f is R integrable on [a, c] and on [c, b] and we have,
$\int_{a}^{b}$$f(x)$ = $\int_{a}^{c}$$f(x)$ + $\int_{c}^{b}$$f(x)$.
This can be shown by taking suitable step functions and partition of $[a, b]$ for this,
Let $P_1$ = {$x_0$, $x_1$,...,$x_m$} be partition of [a, c] and $P_2$ = { $x$$_m$, $x$$_m$$_+$$_1$, ..., $x$$_n$} be partition of [c, b].
Let us define a step function $h$ $:$ [a, b] $\to$ $\mathbb{R}$ defined by $h(x)$ = $c$$_k$ , $\forall$ $x$ $\in$ [$x$$_k$$_-$$_1$, $x$$_k$]
Now, $I$$($$P, $h$)$ = $\sum\limits_{k=1}^n h(x)$ $\Delta$$x$$_k$, $\forall$ $x$ $\in$ [$x$$_k$$_-$$_1$, $x$$_k$]
=$\sum\limits_{k=1}^n c$$_k$ $\Delta$$x$$_k$
=$\sum\limits_{k=1}^m c$$_k$ $\Delta$$x$$_k$ + $\sum\limits_{k=m+1}^n c$$_k$ $\Delta$$x$$_k$
=  $I$$($$P$$_1$, $h$) + $I$$($$P$$_2$, $h$) 
Hence,
$\int_{a}^{b}$$h(x)$ = $\int_{a}^{c}$$h(x)$ + $\int_{c}^{b}$$h(x)$
Now we can replace $h$$($$x$$)$ by $f$$($$x$$)$ since it is definite integral,
So we have,
$\int_{a}^{b}$$f(x)$ = $\int_{a}^{c}$$f(x)$ + $\int_{c}^{b}$$f(x)$.
