# How to show the following function is Riemann Integrable

We have not covered and thus it is not valid to use ideas such as Lebesgue integration, measure, etc. I was given a hint to use either squeeze theorem, or the criterion about if Riemann integrable functions are equal except on a finite set then they have the same integral.

id like to be able to show that the following function is Riemann integrable on $[0,1]$

$$f_{2}(x)=\begin{cases} 1/n &\text{if  \frac{1}{n+1} \lt x \lt \frac{1}{n}} \\ 0 &\text{else} \\ \end{cases}$$

My thoughts,

$f_{2}(x)$ is zero at all rational points with numerator 1,

it is only non zero at irrational for example

$$f_{2}(\pi/6)=\frac{1}{2}$$

$$f_{2}(e/11)=\frac{1}{4}$$

But I dont know how to comine this via comparing functions/squeeze theorem etc.

I already proved that $$f_{1}(x)=\begin{cases} 1/n &\text{if x=\frac{1}{n}} \\ 0 &\text{else} \\ \end{cases}$$ is Riemann integrable on $[0,1]$ with a value of zero. Maybe this could come in some use. I dont need to use it though its just a thought. How else could I go about showing it though? I am overall confused though and looking for help. Thanks

• The set of points where $f$ is not continuous has Lebesgue measure zero, so the function is Riemann integrable. – Svetoslav Feb 6 '16 at 2:24
• $g(1/n)=1$ for all $n$, so $g(1/n)>f(1/n)=1/n$ for $n\ge 2$. We don't really have $f(x)=g(x)$ on the rationals, since $g(x)\ne 0$ for all rational $x\ne 0$, while $f(x)=0$ unless $x$ has the special form $1/n$, in which case it is not equal to $g(x)$ except when $n=1$. However, $f$ and $g$ are equal on the irrationals. I'm not sure, however, how this helps you: Svetoslav's suggestion is better. Alternatively, you can just construct partitions $P_n$ which surround $1/i$ for $i=1,\ldots,n$ by very small intervals: show that for those partitions the upper and lower Riemann sums converge. – ForgotALot Feb 6 '16 at 2:45
• One more thought: there's probably a theorem out there that if $f$ and $g$ differ on only a countable set, and $g$ is Riemann integrable then so is $f$. But if you have this theorem, you don't need Thomaes' function; just take $g=0$. – ForgotALot Feb 6 '16 at 2:48
• Maybe (although it's always hard to say unless one writes out a full answer). One would prove, I suppose, that for any partition the upper sums of $g$ exceed the upper sums of $f$, and so if $g$ has Riemann integral zero, the upper sums go to zero as the partition becomes finer, and so the upper sums of $f$ do also. – ForgotALot Feb 6 '16 at 4:39
• As I said, I think you can "squeeze" the upper sums of $f$ between the upper sums of $g$ and zero. I suppose since $f\ge 0$, the lower sums of $f$ will be squeezed between the upper sums of $f$ and zero, so everything will get duly squeezed, proving that $f$ is Riemann-integrable. – ForgotALot Feb 6 '16 at 6:57

You can start by showing that if $f \colon [a,b] \rightarrow \mathbb{R}$ is a bounded function and $c \in (a,b)$ such that both $f|_{[a,c]}$ and $f|_{[c,b]}$ are Riemann integrable, then $f$ is Riemann integrable and

$$\int_a^b f(x)\, dx = \int_a^c f(x)\, dx + \int_c^b f(x) \, dx.$$

Then, prove that for any $a < x_0 < b$ and $y_0 \in \mathbb{R}$, the function

$$f_{x_0,y_0}(x) := \begin{cases} y_0 & x = x_0 \\ 0 & x \neq x_0 \end{cases}$$

is Riemann integrable on $[a,b]$ with integral zero. The argument will be the same argument as for your $f_1$.

Now, choose some $\frac{1}{2} < a_1 < 1$. On $[a_1,1]$, your function $f$ is just $1 - f_{1,1}(x)$ and thus, is Riemann integrable. On $[\frac{1}{2},a_1]$, your function is just $1 - f_{\frac{1}{2},1}$ and thus is also Riemann integrable. By the result above, $f|_{[\frac{1}{2},1]}$ is Riemann integrable. Continuing this way inductively, you can see that $f|_{[\frac{1}{n},1]}$ is Riemann integrable for all $n \in \mathbb{N}$. Alternatively, if you already know that a function with finitely many discontinuities is Riemann integrable, you can skip all of the above.

Finally, using the definition of the Riemann integrable directly, show that if $f \colon [a,b] \rightarrow \mathbb{R}$ is a bounded function such that $f|_{[a + \frac{1}{n},b]}$ is Riemann integrable for all sufficiently large $n$, then $f$ is Riemann integrable and

$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \int_{a + \frac{1}{n}}^b f(x)\, dx.$$