question
let $\mu, \nu$ be finite borel measures on $[0, \infty)$. If it holds \begin{align} \int_{[0, \infty)} e^{-nx} d\mu = \int_{[0, \infty)} e^{-nx} d\nu \quad {\text{for n} = 1,3,5, \ldots} \end{align} then, $\mu = \nu$ .
my idea I know that
\begin{align} \int_{[0, \infty)} f d\mu = \int_{[0, \infty)} f d\nu \quad {\text{for any} \, f \in C_b[0,\infty)} \end{align} implies $\mu = \nu$ . I tried to check this, but I am not sure how to express continuous bounded $f$ by $e^{-1x},e^{-3x},e^{-5x},\ldots$ .
Is there any other good way?
This is one possible way to do it.
Consider the measures $\mu'$ and $\nu'$ on $[0,1]$ induced by the map $t\mapsto e^{-t}$, that is $$\mu'(A)=\mu(\{t\geq0:e^{-t}\in A\})$$ and similar for $\nu'$.
Then your problem becomes equivalent to showing that $\mu'=\nu'$ provided that $$\int^1_0 u^n\mu'(du)=\int^1_0u^n\nu'(du)$$ for all $n\in\mathbb{Z}$_+ (I believe in your OP, the identity should extend to $n=0$). Then you have the Stone-Weierstrass theorem at your disposal.
The polynomials of odd powers are dense in $\mathcal{A}=\{f\in C([0,1]): f(0)=0\}$ with the sup norm. $\mathcal{A}$ is dense in $L_1(\nu')$ and $L_1(\mu')$.
I hope you can fill in all the details I left should you want to proceed this way.