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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 $g$ ... ....2011-05-18

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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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    My comments were directed more at the OP than at you... Well, I guess we have yet another instance of [Arnold's principle](http://pauli.uni-muenster.de/~munsteg/arnold.html) aka. [Stigler's law of eponymy](http://en.wikipedia.org/wiki/Stigler%27s_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.2011-05-19
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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.