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Thread: Sequences of Step Functions

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    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}$$, which contradicts the fact that $\displaystyle $\displaystyle \lim_{m \rightarrow \infty} \int_{I} s_{m} \leq \int^{a + \delta + 3M}_{a + \delta} s_{N}$$.
    Last edited by ragnar; Jul 21st 2010 at 09:15 PM. Reason: No longer have a question, really.
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