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I have an equality that holds for any $\lambda > 0$ $$ \int\limits_{0}^{\infty}{e^{-\lambda t^{\alpha}}} T(t)dt = \int\limits_{0}^{\infty}e^{-\lambda t^{\alpha}} dg(t), $$ where $\alpha > 0$ is fixed, $T(t)$ is a generalized function on $[0,\infty)$ such that $\int_{0}^{\infty} T(t) dt < \infty$ and $g(t)$ is some non-decreasing function such that $\int_{0}^{\infty} dg(t) < \infty$. Is it true that $T(t) = g'(t)$ in distributional sense? In other words, is $g'(t)$ the only distribution that sovles the equation for $T(t)$?

If $T(t)$ is defined by measure $\mu$ and $g(t)$ defines measure $\nu$ then we can consider push-forward measures $\mu_{*}$ and $\nu_{*}$ under the action of map $t \to t^{\alpha}$. Then the equality gives $$ \int\limits_{0}^{\infty} e^{-\lambda \tau} d\mu_{*}(\tau) = \int\limits_{0}^{\infty}{e^{-\lambda \tau}} d\nu_{*}(\tau) $$ And from that we can derive that $\mu_{*} = \nu_{*}$ and hence $\mu = \nu$, so $g' = T$. Then if there exists another distribution $T$ that solves the equation it can't be represented by measure.

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