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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?

vilma
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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 06 '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 06 '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 06 '14 at 14:54
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    yeah this is why I think something must be wrong. Duh – vilma Jun 06 '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 06 '14 at 18:24
  • @Siminore: That's just about an as unconstructive comment as you can get. Appeals to authority is bad enough when it comes to science in general, but in mathematics it should be considered a capital crime. Nothing stymies creativity more than fear of trying something new. – Benjamin Lindqvist Jun 11 '14 at 06:05
  • @BenjaminLindqvist It is a matter of fact that when a young mathematician "discovers" a trivial proof of a deep theorem, that proof will be wrong with a probability of 99%. Period. I'll add that I made the same mistake so many times, in the past... – Siminore Jun 11 '14 at 09:44
  • More like 99.99999%, but that's entirely beside the point. Would you rather have someone try to prove something, and instead of asking "where is the fault in my logic here?", simply discard the idea because he's probably "wrong". As if being "right" or "wrong" has any sort of relevance in mathematics. – Benjamin Lindqvist Jun 11 '14 at 16:02

3 Answers3

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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 $U(P)-L(P)\ge 4\pi$.

Hagen von Eitzen
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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.

Rene Schipperus
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