Let $a>b>0$, $g\in L^{1}([0,1])$, and let $$f(x):=x^a \int_{0}^{1}e^{-x^b y} g(y)dy,\quad x>0.$$
Claim: $\lim_{x\rightarrow +\infty} f(x)=0$ provided that $y^{-a/b} g(y) \in L^{1}([0,1])$.
Proof:
Let $F(x):=x^a e^{-x^b y}$. Simple calculus shows that $$F(x)\leq c^{c} e^{-c} y^{-c},$$ for each $y>0$, with $c={a}/{b}$.
So, we have $|F(x)g(y)|\leq C y^{-c} |g(y)|$, for every $x>0$. And since the dominating function $y^{-c} |g(y)|$ is integrable on $[0,1]$, we have
$$\lim_{x\rightarrow +\infty} f(x)=\int_{0}^{1} \lim_{x\rightarrow +\infty} x^a e^{-x^b y} g(y)dy= 0$$
because
$$\lim_{x\rightarrow +\infty} x^a e^{-x^b y}=0, \;\;\forall y>0.$$
My questions:
(1) Is the reasoning above correct ?
More importantly,
(2) Is the condition $y^{-a/b} g(y) \in L^{1}([0,1])$ necessary for the limit $\lim_{x\rightarrow +\infty} f(x)$ to exist (as a finite number) ?
Edit: @Ninad Munshi showed me that when $a=b$, the condition $y^{-1} g(y) \in L^{1}([0,1])$ is not necessary for the limit to exist and be finite. Actually, I am interested in the case $a>b>0$.
Your reasoning does indeed work, and you are justified in swapping the limits. The condition, however, is not required for the limit to exist and be finite.
Suppose $a=b$ and $g(0)\neq 0$ but $g\in L^1([0,1])$ (which would mean $y^{-1}g(y)\notin L^1([0,1])$ and would violate the given condition in the problem). Then using the substitution $x^ay = t$ the limit evaluates to
$$\lim_{x\to+\infty}x^a\int_0^1e^{-x^ay}g(y)\:dy = \lim_{x\to+\infty}\int_0^{x^a}e^{-t}g\left(\frac{t}{x^a}\right)\:dt \longrightarrow\int_0^\infty g(0)e^{-t}\:dt = g(0)$$
by dominated convergence.