Prove that $\lim_{\lambda\rightarrow\infty} \frac{1}{\lambda}\int_0^\lambda\int_0^xf(y)\,dy\,dx = \int_0^\infty f(x)\,dx$

Question

Let $f:[0,\infty)$ be Lebesgue-integrable, then prove that

$$\lim_{\lambda\rightarrow\infty} \frac{1}{\lambda}\int_0^\lambda\int_0^xf(y)\,dy\,dx = \int_0^\infty f(x)\,dx$$

This is also known as Cesàro summability of an integral.

I know that I should use the Fubini-Toenlli Theorem to change the order of the integration, but I don't know how to get there to do it.

Answer 1

Since $f \in L^1$, then $\int_0^x |f(y)| dy = M(y)$ is finite, implying that $\int_0^x f(y) dy = K(x)$ is also finite $\forall x\in [0,\infty]$. Note that

$$ \sup_{x \in [0,\infty]}(M(x)) = M(x)=\int_0^{\infty} |f(y)| dy<+\infty $$ Then, $$ \frac{1}{\lambda}\int_0^\lambda \int_0^x |f(y)| dy dx = \frac{1}{\lambda}\int_0^\lambda M(x) dx \leq \frac{1}{\lambda}\int_0^\lambda \sup_x(M(x)) dx = \frac{1}{\lambda}M(x)\lambda = M(x) $$ Since the Lebesgue measure is $\sigma$-finite and $\frac{1}{\lambda}\int_0^\lambda \int_0^x f(y) dy dx$ is finite, we can use Fubini's theorem, therefore $$ \frac{1}{\lambda}\int_0^\lambda \int_0^x f(y) dy dx = \frac{1}{\lambda}\int_0^\lambda \int_0^{+\infty} f(y) [\mathbb I_{[0,x]}(y) ]dy dx = $$ $$ =\frac{1}{\lambda}\int_0^{+\infty} f(y) \int_0^\lambda \mathbb I_{[0,x]}(y) dx dy = \frac{1}{\lambda}\int_0^{+\infty} f(y) \int_0^\lambda \mathbb I_{[y,+\infty]}(x) dx dy= $$ $$ = \frac{1}{\lambda}\int_0^{+\infty} f(y)(\lambda - y)\mathbb I_{[y\leq\lambda]} dy= \int_0^{+\infty} f(y)(1-y/\lambda)\mathbb I_{[0\leq y\leq\lambda]} \ dy $$

Note that $|f(y)| \geq |f(y)(1-y/\lambda)|$, because $|(1-y/\lambda)|\leq 1$. Finally, use the Dominated Convergence Theorem, therefore $$\lim_{\lambda \to +\infty}\int_0^{+\infty} f(y)(1-y/\lambda)\mathbb I_{[0\leq y\leq\lambda]} \ dy= \int_0^{+\infty}\lim_{\lambda \to +\infty} f(y)(1-y/\lambda)\mathbb I_{[0\leq y\leq\lambda]} \ dy = \int_0^{+\infty} f(y) \ dy $$

Answer 2

This is similar to Cesaro's limit, i.e. if $a_n\rightarrow a$ ($a$ a number) then $\frac{1}{n}\sum^n_{k=1}a_k\rightarrow a$. Try to see if you can reproduce the Cesaro arguments for your integral. (Dominated convergence plays a role here)

Let $I=\int^\infty_0 f$ and $J(x)=\int^x_0f$. Given $\varepsilon>0$ let $a>0$ such that $|I-\int^x_0f|<\varepsilon$ for all $x>a$. Then

$$ \frac{1}{\lambda}\int^\lambda_0J(x)\,dx= I +\frac{1}{\lambda}\int^\lambda_0 (J(x)-I)\,dx=I +\frac{1}{\lambda}\int^a_0 (J(x)-I)\,dx + \frac{1}{\lambda}\int^\lambda_a (J(x)-I)\,dx$$

The term $\frac{1}{\lambda}\int^a_0 (J(x)-I)\,dx\xrightarrow{\lambda\rightarrow\infty}0$. The last term is bounded since $$\Big|\frac{1}{\lambda}\int^\lambda_a (J(x)-I)\,dx\Big|\leq \varepsilon\frac{\lambda-a}{\lambda}$$