Proving that a measurable function is integrable

[Crossfader]

1. Homework Statement [/b]

Let f:ℝ\rightarrowℝ be measureable and A_{k}=\left\{x\inℝ:2^{k-1}<\left|f(x)\right|≤2^{k}\right\}, k\in \mathbb{Z}.

Show that f is integrable only if \sum^{∞}_{k=-∞}2^{k}m(A_{k}) < ∞.

Homework Equations

By the definition f is integrable in ℝ if and only if f is measurable and ∫_{ℝ}\left|f\right|<∞.

Now we know that f is measureable, thus we should show that ∫_{E}\left|f\right|≤ \sum^{∞}_{k=-∞}2^{k}m(A_{k}) < ∞.

The Attempt at a Solution

Let f = \sum^{∞}_{k=-∞}f_{k}, we get \left|f \right| = \left| \sum^{∞}_{k=-∞} f_{k} \right| \Rightarrow \int_{ℝ}\left|f\right| = \int_{ℝ}\left|\sum^{∞}_{k=-∞}f_{k}\right|

(?)We also notice that \int_{ℝ}\left|\sum^{∞}_{k=-∞}f_{k}\right| = \sum^{∞}_{k=-∞}\int_{ℝ}\left|f_{k}\right| = 2 \sum^{∞}_{k=0}\int_{ℝ}\left|f_{k}\right| (Beppo Levi's lemma?) (?)

Let I_{k} = [2^{k-1}, 2^{k}]. The length of this interval is \ell(I_{k}) = 2^{k}-2^{k-1}=2^{k}(1-2^{-1}) = (1/2)2^{k}

We get m(A_{k})≤\ell(I_{k}) \Rightarrow \sum^{∞}_{k=-∞}m(A_{k})≤\sum^{∞}_{k=-∞}\ell(I_{k}) = \frac{1}{2}\sum^{∞}_{k=-∞}2^{k}.

And thereby 2\sum^{∞}_{k=-∞}m(A_{k}) = \sum^{∞}_{k=-∞}2m(A_{k}) ≤ \sum^{∞}_{k=-∞}2^{k}.

Does my inference make any sense?

 

[jbunniii]

What is f_k? You didn't define it.

Also, your assertion that

m(A_{k})≤\ell(I_{k})

is nonsense. A_k is a subset of the domain of the function, whereas I_k is a subset of the range. You certainly do not have A_k \subset I_k, which I suspect is the unstated assertion leading you to claim that m(A_{k})≤\ell(I_{k}). E.g. if I put f(x) = 2^{k} for all x, then A_k = \mathbb{R} and thus m(A_k) = \infty.