# Continuous functions are integrable, why the long proof?

I was trying to prove that a continuous function $f:[a,b]\to\Bbb{R}$ is integrable and thought that I came up with an easy solution so I checked the internet and here is a long proof: https://proofwiki.org/wiki/Continuous_Function_is_Riemann_Integrable. This makes me think that something is wrong with my argument:

We know that $f$ attains its maximum and minimum on $[a,b]$, call them $A$ and $B$ resp. Assume that $f$ is not constant so that $A\not=B$ (when $f$ is constant it is trivial to prove that $f$ is integrable). Let $\epsilon>0$. It suffices to find a partition such that $U(P)-L(P)<\epsilon$. This is true for a partition with mesh $<\epsilon/(A-B)$. Isn't it?

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Your choice of partition gives $n\epsilon$ for the value of $U-L$ where $n$ is the number of subintervals determined by your partition, so not smaller than $\epsilon$. –  Santiago Canez Jun 6 '14 at 14:32
Correction: not $n\epsilon$ for the exact value of $U-L$, but rather as a bound for $U-L$. –  Santiago Canez Jun 6 '14 at 14:46
Don't take this personally, but do you really think that so many mathematicians could have missed such a proof, if it were correct? –  Siminore Jun 6 '14 at 14:54
yeah this is why I think something must be wrong. Duh –  vilma Jun 6 '14 at 16:12
The counter arguments to your candidate proof show what you can learn from this: If your proposed bound depends on characteristics of a given function (in this case the upper and lower bound), there's probably a function somewhere that can break your proof. As Rene Schipperus points out, we need a stronger guarantee, i.e. uniform continuity. Lucky for us, every continuous function on a compact set is uniformly continuous. –  rwols Jun 6 '14 at 18:24

## 3 Answers

No, this proof is not correct - there's not a nice connection between the size of the mesh and the difference in $y$-values. Let's look at a specific function, $y = \sqrt x$, on the interval $[0,1]$. Choose $\epsilon = .1$, so the proposed mesh size is $0.1$.

Looking at the interval partitioned as $[0, .1] \cup [.1, .2] \cup ... \cup [.9, 1]$, we have that the difference between the partitions is at least $\sqrt{.1} > .1$ - just look at the very first piece of the partition.

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Consider $f_c\colon[0,2\pi]\to\mathbb R$, $x\mapsto \sin cx$ where $c$ is some constant. For all $c\ge 1$m you will find $A=1$ and $B=-1$, so your proof suggests that $U(P)-L(P)<\epsilon$ as soon as the mesh is $\delta<\frac{\epsilon}{A-B}$ - no matter what $c$ is. But for suitable $c$ (namely $c>\frac{2\pi}\delta$), you will notice that $f$ assumes the values $A$ and $B$ also in each interval of length $\delta$, hence any such partition will sitll have $U(P)-L(P)\ge 4\pi$.

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Use the fact that a continuous function on a closed interval is uniformly continuous to find $\delta$ such that $$|x-y| \leq \delta \rightarrow |f(x)-f(y)| \leq \frac{\epsilon}{b-a}$$ then $\delta$ is the size of the mesh.

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