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- #1

- Apr 13, 2013

- 3,718

Let $f:[a,b] \to \mathbb{R}$ bounded and $c \in (a,b)$.Then $f$ is integrable at $[a,b]$ iff $f$ is integrable at $[a,c]$ and $[c,b]$.In this case,we have $\int_a^b f = \int_a^c f + \int_c^b f$.

The proof for the direction $\Rightarrow$ is like that:

Suppose that $f$ is integral.Let $\epsilon>0$.As $f$ is integrable there is a partition $P=\{a=t_0<t_1<...<t_n=b\}$ of $[a,b]$ such that $U(f,P)-L(f,P)< \epsilon$.

The point $c$ is in an interval $[t_i,t_{i+1}]$.Let's suppose,without loss of generality,that $c$ is not a endpoint of $[t_i,t_{i+1}]$.

So,we have the partitions:

$P_{1}=\{a=t_{0}<...<t_i<c\}$ of $[a,c]$ and $P_2={c<t_{i+1}<...<t_{n}=b}$ of $[c,b]$.

Therefore,$U(f,P_1)-L(f,P_1)\overset{1}{=}$$[U(f,P \cup \{c\})-L(f,P \cup \{c\})]-U(f,P_2)-L(f,P_2)$$ \leq U(f,P \cup \{c\})-L(f,P \cup \{c\}) \leq U(f,P)-L(f,P) < \epsilon$

But..how do we get to $\overset{1}{=}..$?? Isn't it $P=P_1 \cup P_2 \cup \{c\}$ ?