# prove that $\chi_{(0,1)} - \chi_{S}$ is not a limit of increasing step functions.

(This problem is from Alan J. Weir "Lebesgue integration & measure 3.2 exercise 11)

Definition
Let $A_n$ be sequence of connected bounded subsets (interval) of real numbers. Step function is defined to be the finite linear combination of their charateristic functions. $$\psi=c_1\chi_{A_1}+c_2\chi_{A_2}+...+c_n\chi_{A_n}$$ while $c_k\in \mathbb R$ for $k=1,2,...,n$

Let ${I_n}$ be a sequence of open intervals in $(0,1)$ which covers all the rational points in $(0,1)$ and such that $\sum \int \chi_{I_n} \leq \frac 1 2$. Let $S=\bigcup I_n$ and $$f=\chi_{(0,1)} - \chi_S$$

show that there is no increasng sequence of step function $\{\psi_n\}$ such that $\lim\psi_n(x)=f(x)$ almost everywhere. (by means of increasing, $\psi_n(x)\leq \psi_{n+1}(x)$ for all $x$)

I think $$\psi_n = \chi_{(0,1)} - \sum _{k=1} ^n\chi_{I_k}$$ is what author intended. $\int \psi_n$ is decreasing and $\int \psi_n \geq 1-\frac 1 2$. this shows that $\lim\int \psi_n$ converges. so $\lim \psi_n(x)=f(x)$ almost everywhere. However convergence of $\psi_n$ doesn't prove the non-existence of increasing step function which converges to $f$ a.e.
How can I finish the proof?

Suppose such a sequence $\{ \psi_n\}$ exists. Note that $\psi_n \leq 0$ at rational points since $f \leq 0$ at those points. If a step function is non-positive at rational points it is so at all but countable many points. [ The countable many points I am referring to are the end points of teh intervals on which the function is a constant]. It follows that $\int \psi_n \leq 0$ and hence $\int f \leq 0$ which is clearly a contradiction.