Let $f:[0,\infty)\rightarrow\mathbb{R}$ be a positive function s.t. for all $M>0,f \in R([0,M])$. Which of the following statements are true? Problem:
Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a positive function s.t. for all $ M > 0 $ it occurs that $ f $ is integrable on $ [0,M] $. Which of the following statements are true?
A. If $\lim _{x \rightarrow+\infty} f(x)=0$ then $\int_{0}^{\infty} f(x) d x$ exists and is finite.
B. If $\int_{0}^{\infty} f(x) d x$ exists and is finite then $\lim _{x \rightarrow+\infty} f(x)=0$.
C. If $\int_{0}^{\infty} f(x) d x$ exists and is finite then $\int_{0}^{\infty} f\left(x^{2}\right) d x$ exists and is finite.
D. None of the above.
Attempt:
I marked 'D' as the answer, here are my counter examples for A,B,C:
Counter-example for $ A$:
$ f(x)=\begin{cases} 0 &\text{if}\; x \in [0,1] \\\\ \frac{1}{\sqrt{x}} & 1<x \; \end{cases} $
Counter-example for $ B$:  $ f(x)=\begin{cases} 1 &\text{if}\; x \in \mathbb{N} \\\\ 0 & else \; \end{cases} $
Counter example for $ C$: $ C $ is not correct, here's my disprove,
Assume for the sake of contradiction that $ \int_{0}^{\infty} f\left(x^{2}\right) d x $ exists and is finite.
$ \int_{0}^{\infty} f(x^2 ) d x = \{ u = x^2 , \sqrt{u} = x , du = 2x dx = 2\sqrt{u} dx  \} = \frac{1}{2} \cdot \int_{0}^{\infty} \frac{f( u )}{ \sqrt{u} } d u $ .
By the Integral Direct comparison test, if we take $ \int_{0}^{\infty} \frac{\frac{f( u )}{ \sqrt{u} }}{ \frac{1}{ \sqrt{u} }} du  = \frac{1}{2}  \int_{0}^{\infty} f( u ) du$ which converges to a finite, positive value, but $ \int_{0}^{\infty} \frac{1}{ \sqrt{u} } du $ diverges. Hence, $ \frac{1}{2} \cdot \int_{0}^{\infty} \frac{f( u )}{ \sqrt{u} } d u $ diverges, a contradiction.
Was I correct? I'm not sure and I'll appreciate the help!
Note:  Using integral comparison test in the sense of comparing between integrals, like here https://web.njit.edu/~bg263/Lecture notes and supplements/L19.pdf ( and not in the context of sums )
 A: Your examples for A and B are great. Regarding C, your usage of the comparison test is incorrect. When comparing the integrals of $g(x)$ and $h(x)$ you should take the limit of $\frac{g(x)}{h(x)}$, rather than the limit of $\int \frac{g(x)}{h(x)}$. Here if we take your approach then we do not know that $\lim_{x\to\infty} f(x)$ exists (per part B). Moreover, it seems you tried to prove that $\int f(x^2)\,dx$ would never be converging, which is certainly not true, for instance $f(x)=\frac{1}{x^2}$ which becomes $f(x^2)=\frac{1}{x^4}$, both of which have a converging integral on $[1,\infty)$ (and one can zero them out at $[0,1)$ like the trick you used in part A).
Whether or not C is correct depends on the exact meaning of the word "integrable" and this should be checked in your lecture notes. It is customary to let "integrable" refer to having a proper Riemann integral, so that for instance $\frac{1}{\sqrt{x}}$ would not be integrable on $[0,1]$, even though its improper integral does converge on that interval. In case the word "integrable" is to include the case of a converging improper integral then a counterexample to C has been given in the comments. However, assuming that "integrable" refers strictly to proper Riemann integrals, then C is correct since it is given that $f(x)$ is integrable in $[0,M]$ for every $M>0$, so the only problem is at infinity. Intuitively then C should be correct since $f(x^2)$ approaches its asymptote at infinity faster than $f(x)$ does. To prove this rigorously, transform the integral as you have using $u=x^2$. Then
$$\int_{0}^{\infty} f(x^2 )\, dx = \frac{1}{2} \int_{0}^{\infty} \frac{f( u )}{ \sqrt{u} }\, du.$$
Split this into two integrals. The integral over $[0,1]$ converges by comparison with $\int_0^1 \frac{1}{\sqrt{u}}\,du$, since $f(u)$ is bounded in $[0,1]$, so that
$$\frac{f(u)}{\sqrt{u}}\leq \frac{M}{\sqrt{u}}$$
and the integral of the right hand side over $[0,1]$ converges. The integral over $[1,\infty)$ also converges, by Dirichlet's test, since $\int_1^b f(u)\,du$ is bounded as it is convergent, and $\frac{1}{\sqrt{u}}$ is decreasing and tends to $0$ as $u\to\infty$.
