Let $\ell^n: [0,\infty)\to [0,1]$ be right-continuous and increasing functions s.t. $\ell^n(0)=0$. Given $x>0$ and Brownian motion $(B\_t)\_{t\ge 0}$, can we prove
$$\limsup\_{n\to\infty}\mathbb P[\exists s\in [0,t]:~ x+B\_s\le \ell^n(s)]\le \mathbb P[\exists s\in [0,t]:~ x+B\_s\le \limsup\_{n\to\infty}\ell^n(s)],\quad \forall t>0?$$


PS : It appears that the pathwise inequality
$$\limsup\_{n\to\infty} {\bf 1}\_{\{\exists s\in [0,t]: x+B\_s\le \ell^n(s)\}}\le {\bf 1}\_{\{\exists s\in [0,t]: x+B\_s\le \limsup\_{n\to\infty}\ell^n(s)\}}$$
does not hold.



Question concerning an inequality on probabilities of hitting times in a paper
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$This is not quite obvious, and it has hardly anything to do with the reverse Fatou lemma.


Indeed, for all $s\in[0,t]$, let
\begin{equation\*}
 l\_n(s):=\sup\_{m\colon m\ge n}\ell^m(s), 
\end{equation\*}
so that
\begin{equation\*}
 \ell^n(s)\le l\_n(s)\downarrow l(s):=\limsup\_n\ell^n(s). \tag{0}
\end{equation\*}
So,
\begin{align\*}
 &\limsup\_n P(\exists s\in[0,t]\ x+B\_s\le\ell^n(s)) \\
 \le &\limsup\_n P(\exists s\in[0,t]\ x+B\_s\le l\_n(s)) \\ 
 =&\lim\_n P(\exists s\in[0,t]\ x+B\_s\le l\_n(s))=P(A), 
\end{align\*}
where
\begin{align\*}
 A&:=\{\forall n\ \exists s\in[0,t]\ x+B\_s\le l\_n(s)\} \\ 
& =\{\forall n\ge m\ \exists s\in[0,t]\ x+B\_s\le l\_n(s)\}; 
\end{align\*}
here and in what follows, $m$ is any natural number.
So, it is enough to show that
\begin{equation\*}
 P(A)\overset{\text{(?)}}\le P(C), \tag{1}
\end{equation\*}
where
\begin{equation\*}
 C:=\{\exists s\in[0,t]\ x+B\_s\le l(s)\}. 
\end{equation\*}
We shall actually show that
\begin{equation\*}
P(A\setminus C)\overset{\text{(?)}}=0, \tag{2}
\end{equation\*}
which will of course imply (1).


Suppose that event $A$ occurs. For all $n$, let  

\begin{equation\*}
 s\_n:=\inf\{s\in[0,t]\colon x+B\_s\le l\_n(s)\}; 
\end{equation\*}
of course, $s\_n$ is a random variable (r.v., with values in $[0,t]$ on $A$), depending on the random path of the Brownian motion $(B\_t)$.
Moreover, since $l\_n(s)\downarrow l(s)$ for all $s\in[0,t]$, we have
\begin{equation\*}
 s\_n\uparrow s\_\*
\end{equation\*}
for some r.v. $s\_\*$, with values in $[0,t]$ on $A$.


Consider first the case when $A$ occurs and $s\_n<s\_\*$ for all $n$. Then for all $n$ there is some $t\_n\in[s\_n,s\_\*)$ such that $x+B\_{t\_n}\le l\_n(t\_n)$.
Also, $l\_n(s)$ is nondecreasing in $s\in[0,t]$. So, for all $n$, we have $x+B\_{t\_n}\le l\_n(s\_\*)$.
So,
\begin{equation\*}
 x+B\_{s\_\*}=\lim\_n(x+B\_{t\_n})\le\lim\_n l\_n(s\_\*)=l(s\_\*). 
\end{equation\*}
Thus,
\begin{equation\*}
 A\cap\{\forall n\ s\_n<s\_\*\}\subseteq C. \tag{2.5}
\end{equation\*}


If $A$ occurs and $s\_n=s\_\*=t$ for some $n$ (and hence for all large enough $n$), then for such $n$ we have $x+B\_t\le l\_n(t)$ and hence $x+B\_t\le l(t)$. Thus,
\begin{equation\*}
 A\cap\{\exists n\ s\_n=s\_\*=t\}\subseteq C. \tag{2.75}
\end{equation\*}


If $A$ occurs and if $s\_n=s\_\*<t$ for some $n$ (and hence for all large enough $n$) and if $s\_\*$ is a point of discontinuity of the function $l$, then for large enough $n$ we have $x+B\_{s\_\*}=x+B\_{s\_n}\le l\_n(s\_\*+)$, so that $x+B\_{s\_\*}\le l^+(s\_\*)$, where $l^+(s):=\lim\_n l\_n(s+)$.
So,
\begin{equation\*}
x+B\_d\le l^+(d) \tag{3} 
\end{equation\*}
at some point $d\in D$, where $D$ is the set of all points of discontinuity of the nondecreasing function $l$; there are at most countably many such points.  

Note that for all $u\in[0,t)$ and all $s\in(u,t]$ we have $l(s)=\lim\_n l\_n(s)\ge\lim\_n l\_n(u+)=l^+(u)$.
So, $l(s)\ge l^+(u)$ for all $u\in[0,t)$ and all $s\in(u,t]$.
Now suppose that $C$ does not occur, so that $x+B\_s>l(s)$ for all $s\in[0,t]$ and hence $x+B\_s> l^+(d)$ for each $d\in D$ and all $s\in(d,t]$. In view of, say, the (local) law of the iterated logarithm for the Brownian motion, for each $d\in[0,t)$ the event
$\{x+B\_d\le l^+(d),\ x+B\_s>l^+(d)\ \forall s\in(d,t]\}$ has the zero probability.
In view of (3) and because the set $D$ is at most countable, we conclude that
\begin{equation\*}
 P(A\cap\{\exists n\ s\_n=s\_\*<t,s\_\*\in D\}\setminus C)=0. \tag{4}
\end{equation\*}


Suppose finally that $A$ occurs and $s\_n=s\_\*<t$ for some $n$ (and hence for all large enough $n$) and $s\_\*$ is a point of continuity of the function $l$. Take now any real $\ep>0$. Then there is some real $\de>0$ such that $l(s\_\*+\de)\le l(s\_\*)+\ep$. So, for all large enough $n$,
\begin{equation\*}
 x+B\_{s\_\*}=x+B\_{s\_n}\le l\_n(s\_n+\de)=l\_n(s\_\*+\de)\to l(s\_\*+\de)\le l(s\_\*)+\ep. 
\end{equation\*}
Letting now $\ep\downarrow0$, we get $x+B\_{s\_\*}\le l(s\_\*)$. So,
\begin{equation\*}
 A\cap\{\exists n\ s\_n=s\_\*<t,s\_\*\notin D\}\subseteq C. \tag{5}
\end{equation\*}


Collecting (2.5), (2.75), (4), and (5), we confirm (2), as desired.


