The question goes like this:
Let $f:[0,1]\rightarrow \mathbb{R}$ be an integrable (lebesgue), non-negative function. We denote for all $t>0$, the set:
$A_{t}=\{x\in [0,1]\mid t\leq f(x) \leq 2t\}$
Show that:
- We get $g(t,x)\geq \chi_{A_t}(x)$ for all $t\in \mathbb{R}$ and $x\in [0,1]$, when $g(t,x)={f(x) \over t} \cdot \chi_{A_t(x)}$
- $\int_{[0,1]}f \leq \int_{[0,\infty)}{2 \over t}\biggl(\int_{A_t}f(x)dm(x)\biggl)dm(t)$.
Note:
It's easy to show the first question.
How do we show the second one?
Of course we need to use product measure.
I got that $A_t$ is measurable for all ${t \in \mathbb{R}}$ becuase ${f}$ is measurable, and $A_t$ is an intersection of measurable sets: $A_t=\{x\in [0,1]\mid f(x) \geq t\} \cap \{x\in [0,1]\mid f(x) \leq 2t\}$.
Please help!
Best Answer
Note that $t \leq f(x) \leq 2t$ is equivalent to $f(x)/2 \leq t \leq f(x)$. In particular, we have $$f(x) = 2\int_{0}^{\infty} 1_{A_t}(x)~\mathrm dt. $$Using this, we obtain
\begin{align*}\int_{[0,1]} f(x) ~\mathrm dx &= \int_{\mathbb{R}^2}2\cdot\mathbb{1}_{[0,1]}(x)1_{A_{t}}(x)~ \mathrm dt ~\mathrm dx \\ &\leq\int_{\mathbb{R}^2}2\cdot g(x,t)~\mathrm dt ~\mathrm dx \\ &= \int_{[0,\infty)}\frac{2}{t}\left(\int_{A_{t}} f(x)~\mathrm dx\right)~\mathrm dt.\end{align*} As $g(x,t) \geq 0$, exchanging the order of integration is easily justified using Fubini's theorem.