I read a paper and there is a claim that: if t > 0 fixed, then $\int_{0}^{t} \mathrm{e}^{\left(r-(1 / 2) \sigma^{2}\right) u+\sigma B_{u}} \mathrm{~d} u\stackrel{\text { law }}{=} \int_{0}^{t} \mathrm{e}^{\left(r-(1 / 2) \sigma^{2}\right)(t-u)+\sigma\left(B_{t}-B_{u}\right)} \mathrm{d} u $
why they are equal in law? As for my intuition, $B_u \sim B_t-B_u$ seems like time-reversal property? I am not sure whether this intuition is right. But why $u \sim t-u$?
If you substitute $u$ for $t-u$ in the integral and note that both $B_t - B_u$ and $B_{t-u}$ are normally distributed with mean $0$ and variance $t-u$, then the result pops straight out.