# Help on showing a function is Riemann integrable

Problem Let $(a_k)_{k=1}^\infty$ be the sequence of values of $\mathbb{Q}\cap[0,1]$ (which is countable set). Let $g:\mathbb{R}\rightarrow\mathbb{R}$ by $g=\sum_{k=1}^\infty {1 \over k}\cdot \chi_{a_k}$, that is, $g(x)=0$ if $x\notin\mathbb{Q}\cap[0,1]$ and $g(a_k)={1\over k}$ for all $k\in\mathbb{N}$. Show $g$ is Riemann integrable.

(Note: My attempt is below, but I was told that there are many ways to prove this efficiently.)

Attempted Work My main problem is understanding the function $g$. I thought at first glance that it was the harmonic series, which is unbounded and so then not integrable, but I was told to think more carefully, but I'm stumped for some reason. I think it's the way how its defined twice to me: seems contradictory definition. But regardless, I went ahead to try...

Proof so far: Let $g$ be defined as above. For $g(a_k)={1\over k}$ we have that $g$ is bounded with compact support in $(a_k)\subseteq\mathbb{Q}\cap[0,1]$. Now let $g={1 \over k}={1\over k^m}+{k^{m-1}-1\over k^m}$ for $k=2,...,\infty$, where $f_1={1\over k^m}$ is bounded above (by 1) and $f_2={k^{m-1}-1\over k^m}$ is bounded by some $\epsilon_k$>0. Both $f_1$ and $f_2$ have thier support in $(a_k)$. $f_1$ is Riemann integrable (is this circular logic?) and $||f_2||_\infty\leq\epsilon_k$. Hence, by a Theorem about Lower and Upper Riemann Sums provided, $g$ is Riemann integrable. (I know this last part leaves much to be desired)

Lemma provided Basically we were provided a nameless lemma that said if we could show that:

(1) $f_1$ and $f_2$ existed such that $g=f_1+f_2$ with $f_1$ and $f_2$ bounded with support on $(a_k)$ and

(2) $f_1$ is Riemann integrable and $||f_2||_\infty\leq\epsilon$ for all $\epsilon>0$.

Then $g$ is Riemann integrable.

Any tips, advice or help is really appreciated. Thanks so much in advance!

I think it is easier to work from the definition here.

It should be clear that for any partition $\pi$, we have $L(g,\pi) = 0$.

Pick $\epsilon>0$. Choose $N$ such that $\frac{1}{N} < \frac{1}{2} \epsilon$. It should be clear that we can pick a partition $\pi$ of $[0,1]$ such that the points $a_1,...,a_N$ are contained in intervals of total length $\frac{1}{2} \epsilon$.

Let $\{I_k\}$ be the intervals of the partition $\pi$, and let ${\cal I} \subset \{I_k\}$ be the collection of intervals containing any point $a_1,...,a_N$.

Then \begin{eqnarray} U(g,\pi) &=& \sum_k (\sup_{x \in I_k} g(x)) \ l(I_k) \\ &=& \sum_{I \in {\cal I}} (\sup_{x \in I} g(x)) \ l(I) +\sum_{I \in {\cal I}^c} (\sup_{x \in I} g(x)) \ l(I) \\ &\le& \sum_{I \in {\cal I}} (1) \ l(I) + \sum_{I \in {\cal I}^c} (\frac{1}{N} l(I)) \\ & \le& \frac{1}{2} \epsilon + \frac{1}{N} \\ &<& \epsilon \end{eqnarray} It follows that $g$ is integrable and $\int_0^1 g(x)dx = 0$.

Note: You could view the above as writing $g = f_1+f_2$, where $f_1 = \sum_{k=1}^N \frac{1}{k} 1_\{a_k\}$, and $f_2 = g-f_1$. It should be clear that $\|f_2\|_\infty < \frac{1}{2} \epsilon$, however, I think it is simpler to do the above.

• Thanks a lot! I agree that your definition using partitions is probably much simpler. Jun 13 '13 at 7:00