Monotonically increasing $f$ . 
Let $f$ be a monotonically incresing function from $[0, 1]$ into $[0,
 1]$. Which of the following statements is/are true?
$(1)$ $f$ must be continuous at all but finitely many points in $[0, 1]$
$(2)$ $f$ must be continuos at all but contably many points in $[0, 1]$
$(3)$ $f$ must be Riemann integrable 
$(4)$ $f$ must be Lebesgue integrable

I can prove that $(3)$ is true.. and so, $(4)$ is also true. But for the remaining cases?
 A: The second case is true. The reason is that for a monotone increasing function, a discontinuity can only be a jump discontinuity, which means that it removes an interval from the range. The deleted intervals are disjoint because of the monotonicity. Because there are countably many rationals, there can only be countably many such intervals deleted.
But the first case is false, in that there can be countably infinitely many discontinuities. For example, you could have a function which is $1-1/n$ on $[1-1/n,1-1/(n+1))$ for $n=1,2,\dots$ and $1$ at $1$.
A: You can firstly consider the function:
$$f(x)=2^{\lceil \log_2(x)\rceil}$$
which has a discontinuity at every power of $2$ in the interval. This provides a counterexample to $1$.
For $2$, notice that we can sum the lengths of the discontinuities; to be quite formal, notice that wherever $f$ is discontinuous, there is a half-open interval missing from the image $f([0,1])$. The measure of these missing half-open intervals cannot exceed $1$, and every discontinuity must contribute an interval of positive measure to $[0,1]\setminus f([0,1])$. Think about why we can't have uncountably many intervals of positive measure while having their union have finite measure.
