## anonymous one year ago Prove that every simple function f(x) can be written in a form: f(x) = sum from n=1to N of αnχAi(x), sets Ai being mutually disjoint and αn not equals αn0 when n not equals n0, N < ∞.

1. anonymous

$\sum_{n=1}^{N} an XAi(x)$

2. anonymous

Unfortunately my entire response got deleted, but here we go again x_x This isn't a proof, more of just an idea and applying that idea in a basic case. But try thinking of it from the point of view of a step function. As you probably know, a step function can be written as: $\sum_{n=1}^{N}a_{k}\mathcal{X}_{R_{k}}(x)$ where each $$R_{k}$$ is a rectangle. If your rectangles are not disjoint, like in a basic example like this: |dw:1443516783423:dw| Clearly the intervals are not disjoint. But we can always refine the intervals to make them disjoint while still maintaining the value of the step function. (Just for reference, the lebesgue integral of this step function would give a value of 4(3) + 1(2) = 14). Wherever we have overlap, we will just add the values along that interval and rewrite as a single constant times the characteristic. In the picture, the step function can be written as: $$\phi = 4\mathcal{X}_{[0,3]} + \mathcal{X}_{[2,4]}$$ On the overlap, which occurs on [2,3], we add the values, which gives us 5, and rewrite that interval as 5 times the characteristic on that interval. Thus our step function can be rewitten like this: $$\phi = 4\mathcal{X}_{[0,2]} + 5\mathcal{X}_{(2,3]} + \mathcal{X}_{(3,4]}$$ Now we have the step function rewritten such that each (R_{k}\) is disjoint. (And the lebesgue integral gives us the same value 4(2) + 5(1) + 1(1) = 14). See if you can apply that idea to your simple functions. Of course now you're dealing with arbitrary measurable sets, but the idea is the same. How can you rewrite two sets that may overlap as a union of disjoint sets?

3. anonymous

$\sum_{n=1}^{N}a_{n}\mathcal{X}_{R_{n}}$ I meant this of course. Sorry about mixing k in there, not meant to be there of course.