Prove that ${\chi}_{(0,1)}-{\chi}_{S}$ is not a limit of increasing step functions.

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 $\sum \int {\chi}_{{I}_{n}}\le \frac{1}{2}$. 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, ${\psi}_{n}(x)\le {\psi}_{n+1}(x)$ 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 $\int {\psi}_{n}\ge 1-\frac{1}{2}$. 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?