# Let An 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}+cdots+c_n chi_{A_n} while c_k in mathbb{R} for k=1,2,...,n

Prove that ${\chi }_{\left(0,1\right)}-{\chi }_{S}$ 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 $f={\chi }_{\left(0,1\right)}-{\chi }_{S}$ show that there is no increasng sequence of step function $\left\{{\psi }_{n}\right\}$ such that $lim{\psi }_{n}\left(x\right)=f\left(x\right)$ almost everywhere. (by means of increasing, ${\psi }_{n}\left(x\right)\le {\psi }_{n+1}\left(x\right)$ for all x)
I think ${\psi }_{n}={\chi }_{\left(0,1\right)}-\sum _{k=1}^{n}{\chi }_{{I}_{k}}$ 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 $lim{\psi }_{n}\left(x\right)=f\left(x\right)$ 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?
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Hassan Watkins
Explanation:
Suppose such a sequence $\left\{{\psi }_{n}\right\}$ exists. Note that ${\psi }_{n}\le 0$ at rational points since $f\le 0$ 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 $\int {\psi }_{n}\le 0$ and hence $\int f\le 0$ which is clearly a contradiction.