# Thread: Today's integration proof #2

1. ## Today's integration proof #2

Let $f:[a,b]\to\mathbb R$ be a bounded function on $[a,b]$ and let $c\in\mathbb R$ satisfying $a

$f$ is integrable in $[a,b]$ iff, it is in $[a,c]$ and in $[c,b].$ Provided the existence of those integrals, prove that $\int_a^bf=\int_a^cf+\int_c^bf.$

2. Originally Posted by Krizalid
Let $f:[a,b]\to\mathbb R$ be a bounded function on $[a,b]$ and let $c\in\mathbb R$ satisfying $a

$f$ is integrable in $[a,b]$ iff, it is in $[a,c]$ and in $[c,b].$ Provided the existence of those integrals, prove that $\int_a^bf=\int_a^cf+\int_c^bf.$
We will first attempt to prove that $f\in\mathfrak{R}[a,b]\Longleftrightarrow f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]$

First let us prove that if $f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]\implies f\in\mathfrak{R}[a,b]$.

Since $f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]$ we can find partitions $P_1,P_2$ of $[a,b][c,b]$ respectively such that

$U\left(P_1,f\right)-L\left(P_1,f\right)<\frac{\varepsilon}{2}$

and

$U\left(P_2,f\right)-L\left(P_2,f\right)<\frac{\varepsilon}{2}$

Adding the two together gives

$\left(U\left(P_1,f\right)+U\left(P_2,f\right)\righ t)-\left(L\left(P_1,f\right)+L\left(P_2,f\right)\righ t)<\varepsilon\quad{\color{red}(*)}$

Now there must exist parition $[a,b]$ of the form $P=P_1\cup P_2$. Where the result follows since

$U\left(P,f\right)-L\left(P,f\right)<{\color{red}(*)}\quad\blacksquar e$

The proof for $f\in\mathfrak{R}[a,b]\implies f\in\mathfrak{R}[a,c]\wedge f\in\mathfrak{R}[c,b]$ follows by assuming the opposite and using the above.

Now to prove the integral one again consider partitions $P_1,P_2$ of $[a,c][b,c]$ respectively. Now once again there must exist a partition of $[a,b]$ of the form $P=P_1\cup P_2$, where it follows

$L\left(P,f\right)\leqslant L\left(P_1,f\right)+L\left(P_2,f\right)$

Taking the supremum over all partitions of the appropriate intervals gives

$\int_{[a,b]} f~dx\leqslant \int_{[a,c]} f~dx+\int_{[c,b]} f~dx\quad\color{red}(1)$

But it must also be true that

$U\left(P_1,f\right)+U\left(P_2,f\right)\leqslant U\left(P,f\right)$

Taking the infimum over all the partitions of the appropriate intervals gives

$\int_{[a,c]} f~dx+\int_{[c,b]} f~dx\leqslant \int_{[a,b]} f~dx\quad\color{red}(2)$

Combining $\color{red}(1),(2)$ gives

$\int_{[a,c]} f~dx+\int_{[c,b]} f~dx=\int_{[a,b]} f~dx\quad\blacksquare$

Any criticism would be greatly appreciated!