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Does the following hold true $\forall T>0,a>0,c>0$ (in particular for c arbitrarily small):
$P_0(\int_0^T e^{-aB_s}ds<{c})>0$?

This is a minor result which will improve several steps in a couple of proofs in my thesis but although it seems intuitive to me that it is a correct statement, I'm having trouble showing it.

The distribution is actually available in Borodin and Salminen but is of quite a complex form and being that the result I'm looking for is quite weaker than knowing the whole distribution, I'm hoping it can be shown without trying to tackle the distribution function.

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up vote 1 down vote accepted

Yes. Let $M$ be defined by $e^{-aM}=c/(2T)$. Now if $(B_s)$ satisfies $B_s>-1$ for all $s\in [0,c/(2e^a)]$ and $B_s>M$ for all $s\in[c/(2e^a),T]$, then $\int_0^{c/(2e^a)} e^{-aB_s}\,ds\le e^a\cdot c/(2e^a)=c/2$.

Similarly $ \int_{c/(2e^a)}^Te^{-aB_s} \, ds < Te^{-aM} < c/2$. Combining the two, assuming that $(B_s)$ satisfies the above conditions, you see $\int_0^T e^{-aB_s}\,ds < c$. Since the conditions are satisfied with positive probability, the desired conclusion holds.

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