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Suppose that $ f : [a,b] \rightarrow \mathbb{R}$ is Riemann integrable on $[a,b]$ and $g:[a,b] \rightarrow \mathbb{R}$ differs from $f$ at only one point $x_0 \in [a,b]$, that is, $g(x)=f(x)$ for $x \neq x_0$ and $g(x_0) \neq f(x_0)$. Show that $g$ is Riemann integrable on $[a,b]$.

I'm having a little trouble, my thing was that maybe find a partition and look at how it behaves in the partition containing $x_0$

Appreciate any help

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Pick any partition and any set of sample points. You only have two cases: $x_0$ is a sample point or is not. What can you say in each case about the absolute difference between the Riemann sum of $f$ and $g$ ... .... –  N. S. May 18 '11 at 18:46
    
If this is a homework problem please use the [homework] tag! –  Asaf Karagila May 18 '11 at 21:00

2 Answers 2

Here is an unnecessarily slick answer:

There is a famous Lebesgue criterion for Riemann integrability of a function $f: [a,b] \rightarrow \mathbb{R}$ (my colleague Roy Smith informs me that it can actually be found already in the work of Riemann!): it is necessary and sufficient that $f$ be bounded and that its set of discontinuities have (Lebesgue!) measure zero.

Given this: it is an easy exercise to show that modifying a function by changing its values at any finite set $S$ does not change its boundedness/unboundedness, and similarly could only create or destroy discontinuities at $x$ for $x \in S$. So the Lebesgue criterion applies here. (Beware: changing a function at a countable set of values only can change the continuity at every point: I leave it to the reader to supply the canonical example of this.)

Of course one can -- and should -- also show this directly from the definition of Riemann integrability.

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Concerning: Riemann vs. Lebesgue-criterion. That's actually not so surprising. You don't need Lebesgue measure to say what it means for a set to be of Lebesgue measure zero. A set is of measure zero if and only if for each $\varepsilon$ you can cover it with intervals whose lengths add up to no more than $\varepsilon$. That's precisely what you need for proving the criterion you mention. By the way, I learned that result under the name Riemann's criterion in my first year analysis course (in Switzerland). –  t.b. May 18 '11 at 22:55
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In case you're interested, here's the reference: The criterion is contained in section 5 of Riemann's work Über die Darstellbarkeit einer Funktion durch eine trigonometrische Reihe and can be found on page 226 and 227 of his collected works, available e.g. here. It's one of three parts of his celebrated Habilitationsschrift of 1854. –  t.b. May 18 '11 at 23:23
    
@Theo: thanks. Of course I don't doubt Roy, and it is not really contradictory or anything like that, since as you say one hardly needs the notion of a measure to define "measure zero", but still it is interesting since -- in all anglophone sources I have encountered, at least -- this result is invariably attributed to Lebesgue. –  Pete L. Clark May 19 '11 at 3:31
    
My comments were directed more at the OP than at you... Well, I guess we have yet another instance of Arnold's principle aka. Stigler's law of eponymy at work here. More seriously: it may well be that the fact that Lebesgue measure makes conceptual sense out of Riemann's condition (whereas in Riemann it appears more like an arbitrary technical condition) is honoured here. Moreover, it was rather towards the turn of the century that the really pathological functions showed up. –  t.b. May 19 '11 at 9:07

General hint: if you are faced with a problem, first try to understand what makes the problem interesting, and what is the noise; then filter the noise and extract the interesting part.

Let $\delta_z(x) = 1$ for $x = z$ and $\delta_z(x) = 0$ otherwise. If we showed that for any $z$ the integral $\int \delta_z(x) dx$ exists and equals $zero$ (this is the interesting part), we would have: $$g = f + c\delta_z(x)$$ and (this is the noise done by your profesor during the course) $$\int g(x) dx = \int f(x) + c\delta_z(x) dx = \int f(x) dx + c\int \delta_z(x) dx = \int f(x) dx$$

The interesting part is up to you.

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