Subdividing an integral into a sum of integrals over a given interval

["Don't panic!"]

How does one prove the following:
\int^{c}_{a} f\left(x\right)dx = \int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx
where f\left(x\right) is continuous in the interval x\in \left[a, b\right], and differentiable on x\in \left(a, b\right).

My approach was the following:

Given that \int^{b}_{a} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right), where x^{\ast}_{i} \in\left[x_{i},x_{i+1}\right], we have that
\int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) + \lim_{n\rightarrow \infty} \frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)
\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \left[\frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)+\frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)\right] = \lim_{n\rightarrow \infty} \frac{1}{n}\left[\left(b-a\right) +\left(c-b\right) \right]\sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)
\qquad\qquad\qquad\qquad\qquad\quad= \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) = \int^{c}_{a} f\left(x\right)dx

However, I have a feeling that this isn't quite correct?!

 

[micromass]

You have made a notational error:

\int^{b}_{a} f\left(x\right)dx +\int^{c}_{b} f\left(x\right)dx = \lim_{n\rightarrow \infty} \frac{\left(b-a\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right) + \lim_{n\rightarrow \infty} \frac{\left(c-b\right)}{n} \sum^{n}_{i=1} f\left(x^{\ast}_{i}\right)

You have said

\int_a^b f(x)dx = \lim_{n\rightarrow \infty} \frac{b-a}{n} \sum_{i=1}^n f(x^*_i)

which is fine. But then you also do

\int_b^c f(x)dx = \lim_{n\rightarrow \infty} \frac{c-b}{n} \sum_{i=1}^n f(x^*_i)

However, this is wrong. You cannot use the same notation $x^*_i$ in both equations. Indeed, the x_i^* of the integral $\int_a^bf(x)dx$ are not the same $x_i^*$ of the other integral. So you need to denote it as something else such as

\int_b^c f(x)dx = \lim_{n\rightarrow \infty} \frac{c-b}{n} \sum_{i=1}^n f(y^*_i)

Also, your points $x_i^*$ should be depending on $n$, since if $n$ changes, then so do these points. So you should probably call them $x_{i,n}^*$.

 

["Don't panic!"]

Ah ok, thanks for pointing that out. How should I proceed with the proof from that point then?

 

[HallsofIvy]

What you can do is choose a refinement of every partition of [a, c] by adding the point "b". Then you can separate the sum from a to b and the sum from b to c.

 

["Don't panic!"]

Ah ok, so I should start with \int_{a}^{c} f\left(x\right) dx = \lim_{n\rightarrow \infty} \frac{\left(c-a\right)}{n}\sum_{i=1}^{n} f\left(x_{i,n}^{\ast}\right), and then redefine the point x_{i,n}^{\ast} by adding the point "b", such that I can split the sum up on the RHS? (i.e. choose a set of points, x_{i,n}^{\ast} \in\left[ x_{i},x_{i+1} \right] in the interval x\in \left[a,b\right], and then another one, y_{j,n}^{\ast} \in\left[ y_{j},y_{j+1} \right] in the interval x\in \left[b,c\right])