Although I know the following property to be true and use it extensively $$\int_{a}^{b}f(x) dx= \int_{a}^{c}f(x) dx + \int_{c}^{b}f(x) dx$$ I'm trying to prove it and I'm not sure that my proof is correct?!
Let $$g(x) = \left\{\begin{array}{l l} f(x) & \quad x\in (a, c)\\0 & \quad \text{elsewhere}\end{array} \right.$$ and $$h(x) = \left\{\begin{array}{l l} f(x) & \quad x\in[c,b) \\0 & \quad \text{elsewhere}\end{array} \right.$$ Clearly $(a, c) \cup [c, b)=(a, b)$ and $f(x)=g(x) +h(x)$ for $x\in (a, b)$. As such, $$\int_{a}^{b} f(x)dx=\int_{a}^{b} \left(g(x) +h(x)\right)dx= \int_{a}^{b} g(x)dx+\int_{a}^{b} h(x)dx\\ \qquad\qquad\quad\quad = \int_{a}^{c} g(x)dx+ \int_{c}^{b} h(x)dx= \int_{a}^{c} f(x)dx+ \int_{c}^{b} f(x)dx.$$ where we have used the linearity of the integral and that $g(x)=0 $ outside the interval $(a,c)$ and $h(x)=0$ outside the interval $[c, b)$.
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$\begingroup$I may be wrong but it seems as if your answer uses precisely what you are trying to prove when you split $\int_a^b g(x) dx$ into $\int_a^c g(x) dx + \int_c^b 0 dx$.
If would like to be completely rigorous with your proof, what you would like to do is consider the fact that since $f$ is integrable on $(a, b)$ (I'm assuming, otherwise it would not make sense), then for every $\epsilon > 0$, there exists a partition $P$ of $(a, b)$ such that $U(f, P) - L(f, p) < \epsilon$. From there the hint is to define separate partitions on the intervals $(a, c)$ and $[c, b)$, say $P'$ and $P''$, respectively. Then from there you can show that the function on each of those intervals is integrable. From there it is just a matter of showing that your lower and upper sums for $P'$ and $P''$ are also bounded by the original upper and lower sums for the partition $P$.
$\endgroup$ 1 $\begingroup$Your proof is correct, although it is usually more standard to use indicator functions instead of defining new variables. For instance, the proof with indicator functions would be as follows:
For a given set $A$, let $$\chi_A(x) = \left\{\begin{array}{l l} 1 & \quad x\in A\\0 & \quad \text{else}\end{array} \right.$$
Then we see that $\chi_{(a,b)} = \chi_{(a,c]} + \chi_{(c,b)}$. Thus we have \begin{align*} \int\limits_a^b f(x)dx &= \int\limits_\mathbb{R} f(x)\chi_{(a,b)} dx = \int\limits_\mathbb{R} f(x)(\chi_{(a,c]} + \chi_{(c,b)}) dx \\ &= \int\limits_\mathbb{R} f(x)\chi_{(a,c]}dx + \int\limits_\mathbb{R} f(x)\chi_{(c,b)}dx = \int\limits_a^c f(x)dx + \int\limits_c^b f(x)dx\end{align*}
The basic idea of your proof is identical to the one I wrote above.
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