Let each $f_n(x)$ function's graph be diagonal-line segments connecting discrete $x$ points having $f(x)=0\ $ with discrete $x$ points having $f(x)=1$. This creates a jagged sawtooth function curve between $\ x=0\ $ and $\ x=1$. Moreover, $f_n(x)$ equals $\ 1\ $ at $\ x = i/k\ $ for $\ k = 1,2,...,n\ $ and, indexed for each $k$, $\ i = 0, 1,...,k$. Here, each $f_n(x)$ equals $\ 0\ $ only at halfway $x$ values, so-named because they are distributed halfway between the $x$ values.
For example, $f_1(x)=0\ $ when $x=1/2$, because $\ 1/2\ $ is halfway between $\ x=0\ $ and $\ x=1$. Similarly, $f_2(x)=0\ $ at $\ x=1/4\ $ and $\ x=3/4$, because they are halfway between $\ x=0\ $ and $\ x=1/2$, and $\ x=1/2\ $ and $\ x=1$, respectively. Other $f_n$'s have halfway values defined similarly. Finally, each graph of $f_n$ is formed in straight-line segments that connect x-value points (having $y=1$) to its neighboring halfway $x$-value points (having $y=0$).
It is asked to prove $\lim_{n\to\infty}\int f_n=1/2$, even though $f_n$ does not converge almost everywhere.
The first part is trivial, and I can show why $f_n$ converges to $\ 1\ $ at rational numbers. However, I am not sure how to show why $f_n$ does not converge at (a set with nonzero measure of) irrational numbers to complete the proof. Can someone help me here?
So far I have tried bounding the $y$-value change for an irrational number for an increase in $n$, but I am not sure I am on the right track. I think I might need to find a pattern for the gap between rational numbers at a given location for large $n$ values?
(This question is taken from problem 23, page 108, "The Lebesgue Integral for Undergraduates", by William Johnston.)