# Prove that the Riemann integral of this function is zero

Here is my question that I am struggling with:

Assume $a, b \in \mathbb R, a \lt b, c \in [a, b], d \in \mathbb R, d \neq 0$. Consider $f : [a, b] \to \mathbb R$ deﬁned by $$f(x) = \begin{cases} 0, &x \in [a, b] \smallsetminus \{c \}, \\ d, &x = c. \end{cases}$$ Use the deﬁnition of the Riemann integral to show that $\int_{a}^{b} f = 0$.

I know the definition of a Riemann integral has $\|P\| \lt \delta$, $|S(f:\dot P)-L|<\epsilon$, and that $S(f:\dot P) = \sum f(t_{i})(x_{i}-x_{i-1})$, but I'm not sure how to use that to prove that the integral is $0$. Any tips?

Added based on the comments. $P$ is a uniform partition, $P= \{x_i \}_{i=0}^n$, and $\dot P$ is a tagged partition, $\dot P= \{ t_i \}_{i=1}^{n}$.

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What do $P$ and $\dot P$ (i.e., P with a dot on top) denote? –  Srivatsan Dec 3 '11 at 23:21
$P$ is a uniform partition, $P = \{x_{i}\}_{i=0}^{n}$, and $\dot P$ is a tagged partition, $\dot P = \{t_{i}\}_{i=1}^{n}$ –  rioneye Dec 3 '11 at 23:29

You just have to consider the partitions $\dot P$ where $c=t_j$ for some $j$. Otherwise, $f(t_i) = 0$ for any other $t_i \in P$. When $c=t_j$, you just have to make $|x_i - x_{i-1}|$ small enough so $f(t_i)|x_i - x_{i-1}|=d|x_i - x_{i-1}|< \epsilon$.

Try to generalize this question proving that if two functions differ only in a finite (or null-measure, if you know what a null-measure set is) set, then they have the same integral.

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For each possible Riemann sum, either all of its terms are $0$, or else (when one of the $t_i$s happen to equal $c$) exactly one term is nonzero. In the latter case the Riemann sum is $d$ times an interval length. Now when the maximum interval length goes towards zero, then ...

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Without loss of generality $c=a$ then the $S(f:\dot P)= \sum \limits_{i=0}^nf(t_i)|t_i-t_{i+1}|=f(a)|t_i-t_{i+1}|$ now since the partition of the interval is $\delta$-fine and $f(a)=d$ this reads as $d\cdot \delta$ and the Riemann-Integral is defined as the $\lim \limits_{\delta\to 0} \; S(f:\dot{P})= \lim \limits _{\delta\to 0} \;d\cdot\delta=0.$

which is what you wanted to show.

I figured $P$ is a partition of the interval $[a,b]$ and it should be $\delta$-fine but you should be more specific when it comes to defining things.

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