# Thread: Sequences of Step Functions

1. ## Sequences of Step Functions

Suppose $\displaystyle \{ s_{n} \}$ a sequence of increasing step functions which approach a function $\displaystyle f$ on an unbounded interval $\displaystyle I$, and $\displaystyle f(x) \geq 1$ a.e. on $\displaystyle I$. I am asked to show that $\displaystyle \{ \int s_{n} \}$ diverges, where this is the Lebesgue integral.

Here's what I have so far:

[Edit: Nevermind, I think I have it. I'll post what I did below, though, in case anybody is a) curious or b) sees a problem with my solution, since doesn't go into a whole lot of detail.]

Suppose $\displaystyle$\{ \int_{I} s_{n} \}$$converges to M. We may assume w.l.o.g. that \displaystyle I$$ is of the form $\displaystyle$[a, \infty)$$, since we may extend the following argument to the other possibilities: \displaystyle (a, \infty), (-\infty, a], (-\infty, a),$$ and $\displaystyle$(-\infty, \infty)$$. First, we select any \displaystyle \delta > 0$$ and consider the interval $\displaystyle$I_{0} = [a + \delta, a + \delta + 3M$]$. I claim that there is some $N$ such that for all $\displaystyle$x \in I_{0}$,$ $\displaystyle$n \geq N \Rightarrow s_{n}(x) \geq \frac{1}{2}$$. For suppose this is false, then there is some point at which \displaystyle s_{N}(x) < \frac{1}{2}$$ for all $\displaystyle$N$$, contradicting our assumptions. We then know that \displaystyle \displaystyle \frac{1}{2} \cdot 3M \leq \int^{a + \delta + 3M}_{a + \delta} s_{N}$$, which contradicts the fact that $\displaystyle$\displaystyle \lim_{m \rightarrow \infty} \int_{I} s_{m} \leq \int^{a + \delta + 3M}_{a + \delta} s_{N}.