Another large noise limit Note: Here all processes take values in $[0, 1]$.
Let $W$ be a standard one dimensional Brownian motion, and $\sigma > 0$ a constant.
Let $X$ be the solution to the SDE
$$dX_t = \sigma X_t \, dW_t$$
with $X_0 = 1$ a.s.
For every $\varepsilon > 0$, let $A_\varepsilon$ be the event $\{\text{max}_{0 \leq t \leq 1} X_t \geq \frac{1}{\varepsilon}\}$, and let $\mathbb P^\varepsilon$ be the probability measure defined by
$$\mathbb P^\varepsilon (E) =\frac{ \mathbb P(E \cap A_\varepsilon)}{\mathbb P(A_\varepsilon)}.$$
for all events $E$.
Define also for each $\varepsilon > 0$ the process $Y^\varepsilon$ by
$$Y^\varepsilon_t := X_t^{-C(\varepsilon)}.$$
where
$$C(\varepsilon) := \frac{\log \frac{1}{\varepsilon} +\frac{\sigma^2}{2}}{\sigma}.$$
Question: Is it true that
$$\lim_{\varepsilon \to 0} \mathbb E_{\mathbb P^\varepsilon} \left [\int_{0}^1 \lvert Y_t^\varepsilon - e^t \rvert \, dt \right ] = 0?$$
Where $\mathbb E_{\mathbb P^\varepsilon}$ denotes the expectation under $\mathbb P^\varepsilon$.
Remark: It may be useful to adapt the method given by Yuval Peres in the answer here, however I do not know how to deal with the additional integral of $\sigma X_t$ against the Brownian bridge.
 A: $\newcommand{\si}{\sigma}\newcommand{\ep}{\varepsilon}\newcommand\num{\operatorname{num}}\newcommand\den{\operatorname{den}}\newcommand{\R}{\mathbb R}
\newcommand{\vpi}{\varphi}$The conjecture is not true in general.
The limit depends on $\si$. In particular, let us show that the limit in question is, not $0$, but $\infty$ if
\begin{equation*}
    \si>2 + \sqrt3; \tag{-2}\label{-2}
\end{equation*}
(also see the heuristics at the end of this answer).
Indeed, let $P:=\mathbb P$, $P_\ep:=\mathbb P_\ep$, $E_\ep:=\mathbb E_{P_\ep}$,
\begin{equation*}
    m:=\ln\frac1\ep\to\infty,\quad l:=m+\si^2/2,\quad\mu:=-\frac\si2, \quad r:=\frac m\si, 
\end{equation*}
\begin{equation*}
    M_t:=\max_{s\in[0,t]}(W_s+\mu s),
\end{equation*}
\begin{equation*}
    B_t:=\{M_t\ge r\}. 
\end{equation*}
Note that $(X_t)$ is a geometric Brownian motion, so that
\begin{equation*}
    X_t=\exp(\si W_t-\si^2 t/2), 
\end{equation*}
whence
\begin{equation*}
    Y_t:=Y^\ep_t=X_t^{-C(\ep)}=e^{\si l t/2}e^{-l(W_t+\mu t)}  \tag{-1}\label{-1}
\end{equation*}
and
\begin{equation*}
    A_\ep=\{M_1\ge r\}\supseteq B_t; 
\end{equation*}
here and in the sequel, $t\in(0,1)$.
It follows that
\begin{equation*}
    E_\ep Y_t\ge\frac\num\den, \tag{0}\label{0}
\end{equation*}
where
\begin{equation*}
    \num:=Ee^{-l(W_t+\mu t)}1_{B_t},\quad \den:=P(A_\ep). 
\end{equation*}
Formula 1.4.8(1) on p. 256 in Handbook of Brownian Motion - Facts and Formulae, Second Edition, by Borodin and Salminen can be rewritten as
\begin{equation*}
    P(M_t<u,W_t+\mu_t\in dz) \\ 
    =\vpi\Big(\frac{z-\mu t}{\sqrt t}\Big)\frac{dz}{\sqrt t}
    -e^{2\mu u}\vpi\Big(\frac{z-2u-\mu t}{\sqrt t}\Big)\frac{dz}{\sqrt t} 
    \tag{1}\label{1}
\end{equation*}
for $z<u$, where $\vpi$ is the standard normal pdf.
Using \eqref{1} (and noting that $M_t\ge W_t+\mu_t$), one can find
\begin{equation*}
\begin{gathered}
    \num=\int_\R P(M_t\ge r,W_t+\mu_t\in dz)e^{-lz} \\ 
    =
    \frac{1}{2} \left(\text{erf}\left(\frac{\si  t \left(2 m+\si ^2+\si \right)-2 m}{2 \sqrt{2} \si 
   \sqrt{t}}\right)+1\right) \\ 
   \times \exp \left(\frac{1}{8} \left(\frac{4 m^2 (\si  t-4)}{\si }+4 m (\si +1)
   (\si  t-2)+(\si +2) \si ^3 t\right)\right) \\ 
   +\frac{1}{2} e^{\frac{1}{8} t \left(2 m+\si ^2\right)
   (2 m+\si  (\si +2))} \text{erfc}\left(\frac{\si  t \left(2 m+\si ^2+\si \right)+2 m}{2
   \sqrt{2} \si  \sqrt{t}}\right)
\end{gathered}
\end{equation*}
and
\begin{equation*}
\begin{gathered}
    \den=P(M_1\ge r) \\ 
    =
1-\frac{1}{2} e^{-m}
   \left(\text{erfc}\left(\frac{\frac{\si }{2}-\frac{m}{\si }}{\sqrt{2}}\right)-2\right)-\frac{1}{2}
   \text{erfc}\left(-\frac{\frac{m}{\si }+\frac{\si }{2}}{\sqrt{2}}\right).   
\end{gathered}
\end{equation*}
If now $\si>2+\sqrt3$, then the interval $I_\si:=(\frac{4\si-1}{\si^2},\min(1,\frac4\si))$ is nonempty and contained in the interval $(0,1)$. Moreover, for any $\si>2+\sqrt3$ and any $t\in I_\si$, we have $\frac\num\den\to\infty$ (as $m\to\infty$), and hence, by \eqref{0}, $E_\ep Y_t\to\infty$. Thus, by Fatou's lemma, the limit in question is $\infty$. $\quad\Box$

Let me offer two competing heuristics to explain this result:
Heuristic I: The large-deviation effect: The large-deviation event $A_\ep=\{M_1\ge r\}=\{M_1\ge m/\si\}$ (with $m\to\infty$) implies that $W_t\approx mt/\si$. (In this case, this follows, for instance, from the independence of $W_1$ from the Brownian bridge $B_\cdot$, where $B_t:=W_t-tW_1$ for $t\in[0,1]$.) So, on the event $A_\ep$ we have $X_t\approx\exp((m-\si^2/2)t)$ and hence
\begin{equation*}
    Y_t\approx\exp\Big(-\frac{m^2}\si\,(1+o(1))t\Big), \tag{2}\label{2}
\end{equation*}
so that we may expect $\int_0^1 Y_t\,dt$ to be somewhat small on the event $A_\ep$, on the order of $\si/m^2$. The smaller $\si$ is, the more pronounced this effect should be. I think we will indeed have $E_\ep\int_0^1 Y_t\,dt\to0$ if \eqref{-2} does not hold, but I have not checked all the details here.
Heuristic II: The counterbalancing effect of a re-weighting exponential factor: However, if $\si$ is large enough, then the large-deviation effect of Heuristic I may be overshadowed by the factor $e^{-lW_t}$ in the representation of $Y_t$ in \eqref{-1}. Indeed, this exponential factor can be very large for negative values of $W_t$ and negligible for positive values of $W_t$, since $l\sim m\to\infty$. So, even though negative values of $W_t$ are somewhat suppressed by the large-deviation condition $M_1\ge m/\si$, this suppression may be counterbalanced by the re-weighting exponential factor $e^{-lW_t}$, which greatly "favors" negative values of $W_t$. This counterbalancing effect will be more successful when the large-deviation effect is less strong, that is, when the spread/diffusion coefficient $\si$ of the Brownian motion is large enough. In this case,
the conditional expectation of $Y_t$ given $A_\ep$ may resemble much more
the unconditional expectation of $Y_t$, which is
\begin{equation*}
    e^{l^2t/(2+o(1))}=e^{m^2t/(2+o(1))}, 
\end{equation*}
which is very, very large (as $m\to\infty$).
Heuristic II is absent in the previous setting, where we do not have a very influential re-weighting exponential factor, such as the just considered factor $e^{-lW_t}$.
