# Let $I=[0, \infty)$ and $f:I \to R$ a mensurable function such that $|f(t)| \leq \frac{t^\alpha}{1+t}$, where $0 <\alpha <1$.

Let $$I=[0, \infty)$$ and $$f:I \to \mathbb{R}$$ be a mensurable function such that $$|f(t)| \leq \frac{t^\alpha}{1+t}$$, where $$0 <\alpha <1$$. Show that the function $$e^{-tx}f(t)$$ is integrable in $$I \times I$$.

I was trying to show that the module is integrable using Tonelli, but I did not succeed.

Just calculate: $$\int_I{\int_I{e^{-tx}\frac{t^\alpha}{1+t}dx}dt} = \int_I{\frac{t^\alpha}{1+t}\int_I{e^{-tx}dx}dt} = \int_I{\frac{t^\alpha}{1+t}\frac{1}{t}dt} < \int_{0}^{1}{t^{\alpha-1}dt} + \int_{1}^{\infty}{t^{\alpha-2}dt} = \frac{1}{\alpha} + \frac{1}{1-\alpha} < \infty.$$ Note that I used inequality $$1+t>1, 1+t>t$$ respectively for two divided interval $$(0,1)$$ and $$(1,\infty)$$ for $$t.$$