Is the integral $\int_{[0, 1]} a^{-(b-x)} \ d\nu(x)$ small independently of the measure $\nu$?

Question

Let $\nu$ be a positive measure, $b>1$ and let $M\gg 0$ be fixed. Assume $a\ge M$. As part of a more challenging exercise, I am trying to evaluate the integral $$\int_{[0, 1]} a^{-(b-x)} \ d\nu(x).$$ Better saying, I would simply say that this integral is small.

This conclusion seems true to me if, e.g., $\nu$ is the Lebesgue measure. In fact,

$$\int_{[0, 1]} a^{-(b-x)} dx =\frac{a^{-(b-x)}{\log(a)}\Big_{[0, 1]}.$$

My questions are:

1. Is that integral small, independently of the measure $\nu$?

2. If yes, how to prove that?

$\bf{EDIT:}$ Avoiding "drastic" examples like the ones exhibited (rightly) by Sudix in the comments, what about a measure whose support is contained in a compact?

Answer 1

If $x>0$, then $b-x<b$ and $-(b-x)>-b$ and $a^{-(b-x)} < a^{-b}$. So $$ \int_{[0,1]} a^{-(b-x)}\;d\nu(x) \le a^{-b} \nu\left([0,1]\right) $$ with equality when $\nu$ is supported on $\{0\}$. Then we get estimate $M^{-b} \nu\left([0,1]\right)$ as long as $a \ge M$.