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I would like to prove the following theorem:

Theorem If $c \in (a,b)$ and $f$ is integrable on $[a,c]$ and $[c,b]$, then $f$ is integrable on $[a,b]$ and $$\int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f.$$

I will make us of the following lemma:

Lemma For two non-empty sets $A, B$ which are bounded above, $\sup(A+B) = \sup A + \sup B$ and $\inf(A+B) = \inf A + \inf B$.

I would appreciate it if someone could verify my proof below since it's different from the ones I usually see in textbooks.

Proof Let $P_1$ be a partition of $[a,c]$, $P_2$ a partition of $[c,b]$ and $P = P_1 \cup P_2$. Then clearly $P$ is a partition of $[a,b]$, $L(f,P) = L(f,P_1) + L(f,P_2)$ and $U(f,P) = U(f,P_1) + U(f,P_2)$. Then by the lemma, $$\sup\{L(f,P)\} = \sup\{L(f,P_1)\} + \sup\{L(f,P_2)\},$$ and $$\inf\{U(f,P)\} = \inf\{U(f,P_1)\} + \inf\{U(f,P_2)\}. $$ Because $f$ is integrable on the respective intervals, $$\int_{a}^{c}f = \sup\{L(f,P_1)\} = \inf\{U(f,P_1)\} $$ and $$\int_{c}^{b}f = \sup\{L(f,P_2)\} = \inf\{U(f,P_2)\}, $$ so we get that $$\sup\{L(f,P)\} = \inf\{U(f,P)\} = \int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f, $$ showing that $f$ is integrable on $[a,b]$ and that the equality holds.

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  • $\begingroup$ looks right to me $\endgroup$ – user251257 Aug 4 '15 at 8:28
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    $\begingroup$ Your $P\,$ is not arbitrary as a partition of $[a,b]$ because it contains $c \dots$ $\endgroup$ – Tony Piccolo Aug 4 '15 at 11:08
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There is another way of proving the theorem, namely:

Say $I_1 = [a,c]$ and $I_2 = [c,b]$.Then

$$\int_a^c f = \int_a^b f \chi_{I_1} \quad \& \quad \int_c^b f = \int_a^b f\chi_{I_2}$$

So by additivity of the integral

$$\int_a^b f \chi_{I_1} + \int_a^b f\chi_{I_2} = \int_a^b f(\chi_{I_1} + \chi_{I_2}) - \int_c^c f(\chi_{I_1} + \chi_{I_2}) = \int_a^b f(\chi_{I_1} + \chi_{I_2}) = \int_a^b f,$$

where $$\chi_{I} (x) = \begin{cases} 1 & x\in I \\ 0 & otherwise \end{cases}$$

Or

You can use the property that

$$\chi_{I_1 \cup I_2} = \chi_{I_1} + \chi_{I_2} - \chi_{I_1 \cap I_2}.$$

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