# prove that every continuous function is integrable

Can someone tell me whether this is correct thank you!

We know that if a function f is continuous on $[a,b]$, a closed finite interval, then f is uniformly continuous on that interval. This means that if we're given any $\varepsilon > 0$, there exists $\delta > 0$ such that if $x$ and $y$ are any two points in $[a,b]$ with $|x-y| < \delta$, then $|f(x) - f(y)| < \varepsilon$.

So let's say we're given an epsilon. To show that a continuous function f is integrable, we must find a delta such that: For all partitions $\Gamma= \{x_0< \ldots < x_n\}$ of $[a,b]$ with $|\Gamma|:=\max \{x_{i+1} - x_i\} < \delta$ we have $$\mathrm{S}_{\delta} - \mathrm{s}_{\delta} < \varepsilon,$$ where, $$\mathrm{S}_{\delta}:=\inf \Sigma\ M_i(x_{i+1} - x_i)\ \mathrm{and}\ \mathrm{s}_{\delta}:=\inf \Sigma\ m_i(x_{i+1} - x_i),$$

over all partitions $\Gamma$ that satisfies $|\Gamma|<\delta$, with $M_i:=\max f|_{[x_i, x_i+1]}$ and $m_i:=\min f|_{[x_i, x_i+1]}$.

So, to recap, for a given epsilon we must find a delta such that $\mathrm{S}_{\delta} - \mathrm{s}_{\delta} < \varepsilon$. Since $f$ is uniformly continuous on $[a,b]$, we can choose $\delta$ such that $$|x-y| < \delta \ \Rightarrow \ |f(x) - f(y)| < \varepsilon/(b-a).$$

Then, for this $\delta$, we have, for any partition $\Gamma$ with $|\Gamma| <\delta$, that $M_i - m_i < \epsilon/(b-a)$. Therefore,

$$\mathrm{S}_{\delta}-\mathrm{s}_{\delta} < \frac\varepsilon{(b-a)} (b-a) = \varepsilon.$$

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MathJax basic tutorial and quick reference – user31280 Dec 7 '12 at 23:56
If you're asking whether what you did is correct, without using LaTeX and thus making that very difficult to read, then the answer I can give is: yes, it seems to be correct. – DonAntonio Dec 8 '12 at 0:23
This is a classical result in Riemann integration theory. Much better if you look some standard textbooks in real analysis. You may refer Theorem 7.2.6 of the book of Bartle, entitled "Introduction to Real Analysis". – juniven Dec 8 '12 at 0:25
I wrote your text in tex. The proof is correct! – Kelson Vieira Dec 8 '12 at 0:37
@Kelson, you changed a lot of the text as well just adding TeX formatting. I think that is undesirable. – Rahul Dec 8 '12 at 0:57