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Let f a Riemann integrable function (in the proper sense) over [a,b]. Let g be another function that is different from f only in a finite number of points.

How can I show that g is Riemann integrable and that the two integrals (of f and g) over [a,b] are equal?

Thank you!

Gmv93
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    If it doesn't change with 1 point, it doesn't chance with a finite number of points. Take a look here. – StackTD May 05 '17 at 12:59
  • What's more the amount of intervals in your mesh containing your finite number of points is also finite, therefore the impact on the partial sum is at most $2n||f||\delta$ where $\delta$ is the norm of your mesh and $n$ the number of points on which they differ, which obviously goes to 0 as $\delta$ goes to 0 – Astyx May 05 '17 at 13:04

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Let define the functions $\chi_a$ as

$$\chi_a(x) = \begin{cases} 1 & \text{ if } x=a \\ 0 & \text { otherwise }\end{cases}$$

We have that $\chi_a$ is riemann integrable and $\int\chi_a dx = 0$. Now we have that $g-f$ is non-zero in finite many points which means we can write it as a finite sum:

$$g(x)-f(x) = \sum c_a\chi_a$$

A finite sum of rimann integrable functions is Riemann integrable and it's integral is the sum of the integrals of the terms. So $\int g(x)-f(x) dx = 0$ and also $g(x) = f(x) + g(x)-f(x)$ means that $g(x)$ is Riemann integrable and it's integral is:

$$\int g(x) dx = \int f(x) dx + \int(g(x) - f(x))dx = \int f(x) dx + 0$$

skyking
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