For $g(x) = \int_0^\infty \frac{1}{x+y} f(y) \, dy$, Show $m\{ x \in (0,\infty) : g(x) > \lambda \} \le 1/\lambda \cdot \lVert f \rVert_{L^1}$

Question

Q: For $x \in (0, \infty)$ let: \begin{align*} g(x) &= \int_0^\infty \frac{1}{x+y} f(y) \, dy \\ \end{align*} Show that for $f \in L^1(0,\infty)$: \begin{align*} m\{ x \in (0,\infty) : g(x) > \lambda \} &\le \frac{\lVert f \rVert_{L^1}}{\lambda} \\ \end{align*}

My work:

\begin{align*} E_\lambda &= \{ x \in (0,\infty) : g(x) > \lambda \} \\ \end{align*}

By Tchebychev

\begin{align*} m(E_\lambda) &\le \frac{1}{\lambda} \int_0^\infty g(x) \, dx \\ &= \frac{1}{\lambda} \int_0^\infty \int_0^\infty \frac{1}{x+y} f(y) \, dy \, dx \\ \end{align*}

I'm a little stuck here. Is there some integration technique to solve or simplify this integral?

Answer 1

It is $g(x)\leq\displaystyle{\int_0^ \infty\frac{|f(y)|}{x+y}dy\leq\frac{1}{x}\|f\|_1}$. By this, we have that for any $\lambda>0$ it is $\{x\in(0,\infty): g(x)>\lambda\}\subset\{x\in(0,\infty): \frac{\|f\|_1}{x}>\lambda\}=(0,\frac{\|f\|_1}{\lambda})$. Hence $m(\{x: g(x)>\lambda\})\leq m((0,\displaystyle{\frac{\|f\|_1}{\lambda}}))=\displaystyle{\frac{\|f\|_1}{\lambda}}$.