# Fubini-Tonelli exercise

Let $f: \mathbb R_{\geq 0} \to \mathbb R$ measurable such that there is $\alpha \in (0,1)$ with $|f(t)| \leq \dfrac{t^{\alpha}}{1+t}$ for all $t \geq 0$. Let $G: \mathbb R_{\geq 0} \times \mathbb R_{\geq 0} \to \mathbb R$ be defined as $$G(x,t)=e^{-xt}f(t)$$ Prove that $G$ is integrable.

It is sufficient to show that $|G|$ is integrable. By Fubini-Tonelli's theorem, we have $$\int_{(\mathbb R_{\geq 0})^2}|G(x,t)|=\int_{ \mathbb R_{\geq 0}}(\int_{ \mathbb R_{\geq 0}}|f(t)|e^{-xt}dx)dt$$ $$=\int_{ \mathbb R_{\geq 0}}|f(t)|(\int_{ \mathbb R_{\geq 0}} e^{-xt}dx)dt \space(*)$$

If for each $t>0$ I define $f_{n,t}(x)=e^{-xt}\chi _{[0,n]}$, it is clear that $\lim_n f_{n,t}(x)=f(x)$ with $(f_{n,t})_{n}$ an increasing sequence.

The Riemann-Integral $$\int_0^n f_{n,t}(x)dx=\dfrac{1}{t}-\dfrac{e^{-nt}}{t},$$ which coincides with the Lebesgue Integral. So by this and by the monotone convergence theorem, the last expression becomes $$(*)=\int_{\mathbb R_{\geq 0}}|f(t)|\dfrac{1}{t}dt\leq \int_{\mathbb R_{\geq 0}}\dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt$$

I would like to show that this last integral is finite, but I don't know how to prove this. I would appreciate if someone could help me to complete the answer and to correct me if necessary.

## 1 Answer

So consider the integrals $\int_0^1 t^{- \alpha} \textrm{d}t$ if $\alpha \in (0,1)$ and $\int_1^\infty t^{-\beta} \textrm{d}t$ with $\beta \in (1 ,2)$.
Then we must bound $\frac{1}{1+t}$ by $1$ on $t \in [0,1]$ and by $\frac{1}{t}$ for $t \in [1, \infty)$.

• Since $1+t>t$, then $\dfrac{1}{1+t} <\dfrac{1}{t}$, so $(*) \leq \int_0^{\infty} t^{\alpha-2}dt$. Since $0<\alpha<1$, then $-2<\alpha-2<-1$, so $2>2-\alpha>1$, this last inequality shows that the improper integral is finite. From here it follows that the original Lebesgue integral is finite. Is this correct? – user156441 Oct 2 '15 at 8:59
• No, since $t^{-\frac{3}{2}}$ is not integrable as is goes to fast to infinity at $0$. And $t$ is not a constant, so perhaps try $1+t \ge 1$. – Hetebrij Oct 2 '15 at 12:16
• You're right, so maybe I could break up the integral into $\int_0^1 \dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt+\int_1^{\infty} \dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt \leq \int_0^1 t^{\alpha}\dfrac{1}{t}dt+\int_1^{\infty} \dfrac{t^{\alpha}}{t}\dfrac{1}{t}dt$. The last expression equals to $\int_0^1 \dfrac{1}{t^{1-\alpha}}dt+\int_1^{\infty} \dfrac{1}{t^{2-\alpha}}dt$, we have $0<1-\alpha<1, 1<2-\alpha<2$, which means these two integrals are convergent. – user156441 Oct 2 '15 at 16:19
• You're also right, as I was not considering the limit as $t \to \infty$. I will update my answer. – Hetebrij Oct 2 '15 at 16:40