Let $F:[0, \infty)\to [0, \infty)$ be a continuous function. It is given that $$ \int_0^{\infty}f(x) dx<\infty.$$ Determine whether$$ \int_0^{\infty}(f(x))^2 dx<\infty.$$
I think it holds. I write $$[0, \infty)=A\cup B=(x\in [0, \infty): f(x)>1) \cup (x\in [0, \infty): f(x)\leq 1). $$ Clearly $A\cap B=\emptyset$. Let $\mu(A)$ denote measure of a set $A$. Then, I can write $$\int_0^{\infty}f(x) dx=\int_Af(x) dx+\int_Bf(x) dx$$ We can easily see that $\mu(A)<\infty$ otherwise $$\int_A f(x) dx \geq \int_A 1\ dx=\mu(A)=\infty,$$ which is a contradiction as $f$ is non-negative and $\int_0^{\infty}f(x) dx<\infty.$ Further, we can also write $$[0, \infty)=A'\cup B'=(x\in [0, \infty): (f(x))^2>1) \cup (x\in [0, \infty): (f(x))^2\leq 1). $$ Also, we know that if $f(x)>1$, then $(f(x))^2>1$. This means $A=A'$ and $B=B'$. Consequently, we have $$\int_0^{\infty}(f(x))^2 dx=\int_A (f(x))^2 dx+\int_B (f(x))^2 dx$$ $$\leq \int_A (f(x))^2 dx+\int_B f(x) dx$$ as $f^2(x)\leq f(x)$ on $B$. Also, as measure of $A$ is finite and $f$ is continuous, we have $\int_A (f(x))^2 dx<\infty.$ Therefore, we have $$\int_0^{\infty}(f(x))^2 dx<\infty\ \ \text{as}\ \ \int_B f(x) dx<\infty.$$
Is my approach correct? Or I am missing something?
Your mistake: 'as measure of $A$ is finite and $f$ is continuous, we have $\int_A (f(x))^2 dx<\infty$': This is false. $f$ need not be bounded on $A$.
This is false. Let $f(x)=n^{4}(x-n)$ for $n \le x \le n+\frac 1 {n^{3}}$, $f(x)=n^{4}(n+\frac 2 {n^{3}}-x))$ for $ n+\frac 1 {n^{3}} \le x \le n+\frac 2 {n^{3}} $, $n=1,2,3,..$ This meets all your requirements, $\int_0^{\infty} f(x)dx<\infty$ and $\int_0^{\infty} [f(x)]^{2}dx=\infty$. [Use convergence of $\sum \frac 1 {n^{2}}$ and divergence of $\sum \frac 1 n$].