"If $g(x)$ is a positive continuous function on $(0,1)$ and $\int_0^1 g(x) \, dx$ converges. Then $\int_0^1 \sqrt{g(x)}$ also converges."
"If $f(x) \ge 0$ on $[0,\infty)$ and $\int_0^\infty f(x) \, dx$ converges, Then $\lim_{x\to\infty} f(x) = 0$"
Because of $\sqrt{g(x)}\ge {g(x)}$ for $0 \le g(x) \le 1$ , I think statement "1" is false.
And I think statement "2" is true.
Am I right?
Well, $\sqrt{g}$ is bounded above by $\max\{g,1\}$, whose improper integral exists finitely, as you would observe. In fact, speaking more generally, using monotone convergence theorem it can be shown that such a non-negative function must be Lebesgue integrable, and this integral must coincide with the improper Riemann integral. You can, if you please, show that for a finite measure space $(X,B,\mu)$, we always have $L^2(X)\subset L^1(X)$. Kindly ignore this in case you haven't done measure theory yet. And you haven't clearly written the second part of your question, so I would take it that you mean whether or not convergence of the improper integral $\int_0^\infty f$ implies the limit of $f$ is zero as $x\rightarrow\infty$. This is false,because you must realize, while integrating you are essentially computing an area, so modifications of a function on sufficiently "small" sets(like the countable ones, if you please) does not change the value of the integral. With this in mind, you may define $f$ as $f(n)=n$ for all $n$ in $\mathbb{N}$, and zero elsewhere. This should suffice for a counterexample.