Answer to Question #140414 in Calculus for Mathew

Question

If f is continuous on [a,b]t hen \underline_{\int_a^b}f(x)dx=\overline{\int_a^b}f(x)dx

Answer 1

Solution

Since f(x) is continuous on [a,b] then it is uniformly continuous on the interval.

Let $\epsilon>0$ by uniform continuity there exist a $\delta>0$ such that $|f(x)-f(y)|< \epsilon$ whenever $|x-y|<\delta$ such that $x,y\in [a,b]$

We now divide the interval [a,b] into partitions $J_1,J_2,\dots,J_n$ each of lengths $\frac{b-a}{N}$

By definition

$\underline{\int_a^bf(x)}dx\leq \sum_{k=1}^N\left(\inf_{x \in J_k}f(x)\right)|J_k|$

$\overline{\int_a^bf(x)}dx\leq \sum_{k=1}^N\left(\sup_{x \in J_k}f(x)\right)|J_k|$

$\overline{\int_a^bf(x)}dx-\underline{\int_a^bf(x)}dx\leq \sum_{k=1}^N\left(\sup_{x \in J_k}f(x)-\inf_{x \in J_k}f(x)\right)|J_k|$ ...... (1)

However we have

$|f(x)-f(y)|<\epsilon$ $\forall x,y \in J_k$ ,since $|J_k|=\frac{b-a}{N}$ ,In particular we have

$f(x)<f(y)+\epsilon$ $\forall x,y\in J_k$

Taking suprema in x we obtain

$\sup_{x\in J_k}f(x) \leq f(y) + \epsilon$ $\forall y\in J_k$

And taking infima in y we obtain

$\sup_{x\in J_k}f(x) \leq \inf_{y\in J_k} f(y) + \epsilon$

Inserting this into (1) we have

$\overline{\int_a^bf(x)}dx-\underline{\int_a^bf(x)}dx\leq \sum_{k=1}^N\epsilon|J_k|$

$\overline{\int_a^bf(x)}dx-\underline{\int_a^bf(x)}dx\leq(b-a)\epsilon$

But $\epsilon >0$ is arbitrary and (b-a) is fixed. Thus $\overline{\int_a^bf(x)}dx-\underline{\int_a^bf(x)}dx$ cannot be positive, but $\overline{\int_a^bf(x)}dx\geq\underline{\int_a^bf(x)}dx$ which implies $\overline{\int_a^bf(x)}dx-\underline{\int_a^bf(x)}dx=0$ which complete the proof.