Proving that a measurable function is integrable

1. Jul 9, 2012

1. The problem statement, all variables and given/known data[/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 $\sum^{∞}_{k=-∞}2^{k}m(A_{k}) < ∞$.

2. Relevant 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 $∫_{E}\left|f\right|≤ \sum^{∞}_{k=-∞}2^{k}m(A_{k}) < ∞$.

3. 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?

2. Jul 9, 2012

jbunniii

What is $f_k$? You didn't define it.

$$m(A_{k})≤\ell(I_{k})$$
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 $m(A_k) = \infty$.