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Let $f \colon [a,b] \to \mathbb R$ bounded, such that $f(x) = 0$ for every $x \in [a,b]$ except in a set $J$ of measure zero. When we say that a set $J$ has measure zero, if given any $\varepsilon > 0$ there exist a countable collection of open intervals $( a_n ,b_n)$ such that $J \subset \bigcup\limits_{n \in {\Bbb N}} (a_n ,b_n)$ and $\sum \limits_{n \in {\Bbb N}} (b_n - a_n) < \varepsilon$. Prove that in the Riemann sense (I don't know any other sense of integrals) the integral exist and $$ \int_a^b f (x) dx = 0. $$

The existence is easy, but how can I prove the equality? Help me with this please. Don't use Lebesgue integrals, because I can't use it in this exercise. It's from a real analysis course. Thanks!

I know that this result it's more general, Instead of putting $f(x) = 0$, I can put any Riemann integrable function, but the general case, comes off as trivial corollary of this. So why try to prove this.

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    For general knowledge, Lebesgue integrals are considered to be a very (very) main theme in real analysis.2011-12-01
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    If $f(x) = \begin{cases} 1 & \text{if }x\text{ is rational}, \\ 0 & \text{if }x\text{ is irrational}, \end{cases}$ then that seems to satisfy all of the hypotheses even though $f$ is not Riemann-integrable.2011-12-01

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