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Let $A=[-a,a]$, with $a>2$. I have a function $g(x)$ continuous in $A$ such that $$\lim_{s\rightarrow\infty} \int_A x^{2s}f(x)dx = \int_A x^{2s}g(x)dx.$$ Can I claim that $f(x) = g(x)$ for $x\in A$ ?

What would be the reason for that answer.

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It is unclear what you mean by $f(x) \to g(x)$ since there is no limit being taken there; perhaps you mean $f(x) = g(x)$ instead?

In this case, the answer is no. Consider when $a < 1$. then $x^{2s} \to 0$ uniformly on $[-a,a]$ as $s \to \infty$. Thus for any bounded $f$ we see $x^{2s}f(x) \to 0$ uniformly as well and so $$\int_A x^{2s} f(x) \, dx \to 0 = \int_A x^{2s}g(x) \, dx$$ where $g(x)$ is identically zero. But it is certainly not that case that any bounded function $f$ is the zero function.

However, we can say that if $$\int_A x^{n} f(x) dx = \int_A x^n g(x) dx$$ for all $n \in \mathbb N$, then $f(x) = g(x)$ for all $x \in A$. This follows from the Stone-Weierstrass theorem.

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    I updated the question – 2017-01-17
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    In my case a>1, so the integral never goes to zero. I am still not sure if that is enough to guarantee f=g – 2017-01-17
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    It certainly isn't. Just choose any continuous $f$ such that $0 \le f \le 1$, $f = 0$ for $\lvert x \rvert > 1$ and $f = 1$ for $\lvert x\rvert < 1/2$ (you can easily draw a picture of such a function). Then we've essentially reduced to the case when $a < 1$, so $$\int_A x^{2s} f(x) \, dx \to 0$$ as $s \to \infty$ but $f$ is not identically zero. – 2017-01-17
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    I apologise for making this confusion. I rephrased the question. – 2017-01-17