# Help in proving $f \circ \phi \in \mathcal L^1(\lambda) \iff \int_0^{\infty} \frac {f(x)}{\sqrt x} \lambda (dx) < \infty$

Consider the measure space $(\mathbb R, \mathcal B(\mathbb R), \lambda)$ and let $\phi: \mathbb R \rightarrow \mathbb R$ be given by $\phi(x) = x^2$.

I want to show that for $f \in \mathcal M(\mathcal B(\mathbb R))$ we have $f \circ \phi \in \mathcal L^1(\lambda) \iff \int_0^{\infty} \frac {|f(x)|}{\sqrt x} \lambda (dx) < \infty$:

I've computed that the measure $\lambda \circ \phi^{-1}$ is given by $t \mapsto t^{-1/2} 1_{(0, \infty)}(t)$ (density). Should I use the transformation theorem ? I'm out of ideas how to proceed.

Also, I want to prove that $\int_{\mathbb R} f \circ \phi \ \lambda(dx) = \int_0^{\infty} \frac {f(x)} { \sqrt x} \lambda(dx)$, but this can be done by applying transformation with respect to measure ?

• You're missing an absolute value on $f(x)$ in the $\iff$. – Stefan Hansen Dec 2 '14 at 11:55
• You are right, thanks. – Shuzheng Dec 2 '14 at 11:56
• I've solved the second question, however the first I still don't have a clue. – Shuzheng Dec 2 '14 at 11:58

Since $\lambda\circ\phi^{-1}$ has density $t\mapsto t^{-1/2}\mathbf{1}_{(0,\infty)}(t)$ with respect to $\lambda$ we have $$\int_\mathbb{R} \mathbf{1}_B\circ \phi\,\mathrm d\lambda=\int_\mathbb{R}\mathbf{1}_{\phi^{-1}(B)}\,\mathrm d\lambda=\lambda(\phi^{-1}(B))=\int_0^\infty\frac{\mathbf{1}_B(x)}{\sqrt{x}}\,\lambda(\mathrm dx)$$ for all $B\in\mathcal{B}(\mathbb{R})$. That is, we have $$\int_\mathbb{R} f\circ\phi\,\mathrm d\lambda=\int_0^\infty \frac{f(x)}{\sqrt{x}}\,\lambda(\mathrm dx)\tag{1}$$ for all $f$ of the form $f=\mathbf{1}_B$ with $B\in\mathcal{B}(\mathbb{R})$. The following standard argument shows that $(1)$ actually holds for all non-negative, measurable $f$:
• By linearity of the integral, we have that $(1)$ holds for all simple functions.
• For a non-negative, measurable $f$, we may find an increasing sequence of non-negative, simple functions $(f_n)$ such that $f_n\uparrow f$ and hence by the monotone convergence theorem we have that $(1)$ holds for $f$.
For a general measurable $f$ we have \begin{align} f\circ\phi\in\mathcal{L}^1(\lambda)&\iff \int_\mathbb{R}|f\circ\phi|\,\mathrm d\lambda<\infty\iff\int_\mathbb{R} |f|\circ\phi\,\mathrm d\lambda<\infty\\ &\iff\int_0^\infty\frac{|f(x)|}{\sqrt{x}}<\infty. \end{align} Now, to show that $(1)$ holds any $f$ such that $f\circ\phi\in\mathcal{L}^1(\lambda)$ you simply use the decomposition $f=f^+-f^-$ into its positive and negative part, and use $(1)$ on both $f^+$ and $f^-$.