# $f(x)$ is Riemann integrable $\Rightarrow$ $\frac{1}{1 + f^2(x)}$ is Riemann integrable

Let f(x) be Riemann integrable on [a,b]. Then

1. there exist $\lim_{x \rightarrow a+0} f(x)$ and $\lim_{x \rightarrow b-0} f(x)$

2. f(x) has only removable or jump discontinuities.

3. The set of discontinuities is finite (or empty)

4. $\frac{1}{1 + f^2(x)}$ is Riemann integrable on [a,b]

Here is what I know so far. Integrability is equivalent to being bounded and continuous almost everywhere. Using the latter, it's obvious that (3) is false.

The (2) is false because f(x) need not to have a discontinuity.

How to dispove (1) and prove (4)?

• (2) does not say that f actually has to have a discontinuity. But if there is, this dicontinuity has to be either removable or a jump. Apr 20, 2015 at 8:20
• In that case, (2) is true. But according to the authors of the problem (2) is false. So, I believe, they meant "has" as "actually having", not as a possibility. Apr 20, 2015 at 8:23
• For (1), take the standard function $\sin (1/x)$. Thus is obviously Riemann integrable, but has no limit for $x \to 0$. Apr 20, 2015 at 8:57

The example $f(x):=1$ when $x={1\over n}$ $(n\in{\mathbb N}_{\geq1})$, and $:=0$ otherwise in $[0,1]$ disproves 1.–3.
The function $\phi(u):={1\over 1+u^2}$ satisfies $\bigl|\phi'(u)\bigr|\leq{1\over2}$ and therefore is Lipschitz-continuous with Lipschitz constant ${1\over2}$. It follows that the function $g:=\phi\circ f$ will pass any "fluctuation test" that $f$ passes. This implies that $$g(x):={1\over 1+ f^2(x)}$$ is Riemann integrable.
If $f:[a,b]\to [c,d]$ is Riemann integrable, and $g$ is continuous on $[c,d],$ then $g\circ f$ is Riemann integrable 0n $[a,b].$ In 4. we have $g(x) = 1/(1+x^2),$ which is continuous everywhere. So 4. holds.