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Suppose $A = \{a_n\}_1^\infty$ is a sequence of real numbers converging to $0$, and $B = \{b_1, b_2, b_3,...\}$ is a countably infinite subset of the interval $[0,1]$. Consider the function $f$ on $[0,1]$ where \begin{equation*} f(x) = \begin{cases} \hfill a_n \hfill & \text{ if $x = b_n$} \\ \hfill 0 \hfill & \text{ if $x \notin B$} \\ \end{cases} \end{equation*} Prove $f$ is Riemann integrable and determine $\int_0^1 f(x)dx$.

I've seen explanations of this for Thomae's function, which is very similar. In this case, since we are told $B$ is a countable set (and $B$ is the set of discontinuities) we have by Lebesgue's Criterion that $f$ is Riemann integrable.

My intuition tells me that the value of the integral should be zero (the same as the integral of Thomae's function), but I'm unsure how to formally prove this.

Calvin Khor
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user20354139
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    To compute the integral of $f$, use $0\leqslant f\leqslant f_n+\epsilon_n$ where $f(b_k)=a_k$ for every $k\leqslant n$, $f_n(x)=0$ otherwise, and $\epsilon_n=\max{a_k\mid k>n}$ and use that $f_n$ has integral $0$ and that $\epsilon_n\to0$. – Did Mar 17 '17 at 06:52
  • *Typo: ...where $f_n(b_k)=a_k$ etc. – Did Mar 17 '17 at 07:15
  • So for example if $A = {4, 2, 1, 1/2, 1/4, 1/8, ... }$ and $B = {0.05, 0.3, 0.1, 0.7, 0.5, 0.4, ... }$ and we fix $n = 3$ then $f_n$ plots $(0.05, 4), (0.3, 2), (0.1, 1/2)$ and zeros everywhere else. Correct? So that breaks the problem down into looking at the integral over three discontinuities as opposed to infinitely many. Then we can look at what happens as $n \to \infty$. How do we say that $\int_0^1 f_n(x)dx = 0$? – user20354139 Mar 17 '17 at 07:31
  • This approach assumes one knows how to prove that any function which is zero everywhere except at a finite number of points is integrable with integral zero. Do you? If you do not, this is the question you should be asking, not the one above. – Did Mar 17 '17 at 09:07
  • Side note: It is irrelevant if $a_n \to 0$, as long as it remains bounded. – PierreCarre Sep 03 '21 at 08:38

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This follows from the theorem:

If $f_n$ are Riemann integrable functions on $[a,b]$ and $f_n \to f$ uniformly, then $f$ is Riemann integrable and $$ \int_a^b f(x) dx = \lim_{n\to \infty} \int_a^b f_n(x)dx.$$

In this case we choose $$f_n =\begin{cases} a_i &\text{if } x = b_i, i=1, 2, \cdots, n \\0 &\text{otherwise.}\end{cases}$$

This $a_n\to 0$ is used to prove that $f_n \to f$ uniformly. Since $$\int_0^1 f_n(x) = 0$$

for all $n$, we have also $$\int_0^1 f(x) dx = 0.$$

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  • Does $B = {b_1, b_2, b_3,...}$ have to be a subset of $[0,1]$ in increasing order? Or does that not matter? If it is then as $x$ increases obviously $f(x)$ decreases to zero, but if not then it seems $f$ could behave really oddly. – user20354139 Mar 19 '17 at 18:23
  • $B$ does not have to be in increasing order. @antsatsui –  Mar 21 '17 at 23:53