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

Question

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)?

Answer 1

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.

Answer 2

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.