Convergence of Brownian integral

I guess that the indication is to mimick the sandwiching argument, starting from something like the fact that the random variable $X_t$ you are interested in is such that $$ A_t^{1/\sqrt{t}}\mathrm e^{(M_t-1)/\sqrt{t}}\le X_t\le t^{1/\sqrt{t}}\mathrm e^{M_t/\sqrt{t}}, $$ where $M_t=\max\{B_s\mid 0\le s\le t\}$ and $A_t=\displaystyle\int\limits_0^t[B_s\ge M_t-1]\mathrm d s$.

In the RHS, $t^{1/\sqrt{t}}\to1$ and $M_t/\sqrt{t}$ equals $M_1$ in distribution.

In the LHS, $(M_t-1)/\sqrt{t}$ converges to $M_1$ in distribution. If $\log A_t=o(\sqrt{t})$ almost surely (a fact which should follow from Williams's decomposition theorems and/or a Ray-Knight theorem for the local time of $B$), one sees that $X_t$ converges in distribution to $\mathrm e^{M_1}$.

This is essentially a "Laplace Method" argument that allows you to conclude that :

$\left(\int_0^te^{B_s}ds\right)^\frac{1}{\sqrt{t}}$ converges in law to $exp(S_1)=e^{\sup_{s\leq 1}\textit{B}_s}$.

You have the following equalities (sometime only in law):

$$(\int_0^te^{B_s}ds)^\frac{1}{\sqrt{t}}= (t\int_{0}^{1}e^{\sqrt{t}\textit{B}_s}ds)^\frac{1}{\sqrt{t}}=t^\frac{1}{\sqrt{t}}.||e^{\textit{B}_.}||_{L^\sqrt{t}[0,1]} $$

Laplace's Method allows you to conclude that $\lim_{t\to \infty}||e^{\textit{B}_.}||_{L^\sqrt{t}[0,1]}=||e^{\textit{B}_.}||_{L^\infty[0,1]}=e^{\sup_{s\leq 1}\textit{B}_s}$

The only thing you need to apply Laplace Method is that the function is $L^{\infty}$ which is true for a.s. for the paths of $e^{\textit{B}_.}$ over the interval $[0,1]$

Regards

Edit proof of Laplace Method : Let's take a positive $f\in L_\mu^{\infty}(X)$, then first we have for any $p>1$ :
$\|f\|_{L^p}\leq \|f\|_{L^{\infty}}\cdot (\mu(X))^{1/p}$ (here $\mu(X)=1$ but this works for finite measures aswell)

Which gives half of the result.

Second, let's fix $\epsilon>0$, and note $A_\epsilon=\{x \in X s.t.|f(x)|\geq \|f\|_{L^{\infty}}-\epsilon\}$, and such that $\mu(A_{\epsilon})>0$ (there exists such an $\epsilon$ and for every values inferior to this value, the corresponding set has non zero measure, this can be seen by absurd)

Then we have:

$\int_{X}{|f(s)|}^p d\mu(s)=\int_{A_{\epsilon}}{|f(s)|}^p d\mu(s)+\int_{A_{\epsilon}^c}{|f(s)|}^p d\mu(s) \geq (\|f\|_{L^{\infty}}-\epsilon)^p \cdot \mu(A_{\epsilon}) $

Taking the p-th root we get :
$(\|f\|_{L^{\infty}} - \epsilon)\cdot (\mu(A_{\epsilon}))^{1/p} \leq \|f\|_{L^p}$

Letting $p$ going to $\infty$ (note that $\epsilon$ is still fixed) we get :

$$ \lim_{p \rightarrow +\infty}(\|f\|_{L^{\infty}} - \epsilon)\cdot (\mu(A_{\epsilon}))^{1/p} = \|f\|_{L^{\infty}} - \epsilon \leq \lim_{p \rightarrow +\infty} \|f\|_{L^p} (*)$$

As this is true for any $\epsilon>0$ we get the desired result.

$(*)1\ge\mu(A_{\epsilon})>0$