limsup of random variables coverges in distribution I came across the following question when I was studying Theorem 4.1.2 on Durrett's book: Probability Theory and Examples, 4th edition
Given $X_i$ i.i.d., and let $S_n:=X_1+\cdots+X_n$ to be the random walk. He shows that $\limsup S_n=c$ a.s. for some $c\in [-\infty,\infty]$.
Edited on 20 Dec: Our goal is to show that $c=c-X_1$ a.s.
I deleted my previous thoughts on the problem since I figured out a solution myself. 
 A: I figured out a proof myself. Please point out if there is any problems.
Let $S_n':=S_{n+1}-X_1$. Then $S_n'$ has the same distribution as $S_n$. Now $\limsup_n S_n=c$ a.s., and hence in particular, $\lim_{n\to\infty}\sup_{k\geq n}S_k=c$ in distribution. By the previous argument, if we have shown that $\sup_{k\geq n}S_k'$ and $\sup_{k\geq n}S_k$ have the same distribution, then $\lim_{n\to\infty}\sup_{k\geq n}S_k'=c$ in distribution as well.
To show this, we consider two terms first:
\begin{align}
 P(\sup\{S_k,S_{k+1}\}\leq t)
 &=P(S_k\leq t, S_{k+1}\leq t)\\
    &=P(S_k\leq t,S_k+X_{k+1}\leq t)\\
    &=\int_{-\infty}^{t}\int_{-\infty}^{t-x}d(\mu_{X_{k+1}}(y)\times \mu_{S_k}(x))\\
    (\text{By independence}) &=\int_{-\infty}^{t}\int_{-\infty}^{t-x}d\mu_{X_{k+1}}(y)d\mu_{S_k}(x)\\
 \end{align}
    Since $X_{k+1}$ has the same distribution as $X_{k+2}$ and $S_k$ has the same distribution as $S'_k$, we can rewrite the above as
    \begin{align}
 &\int_{-\infty}^{t}\int_{-\infty}^{t-x}d\mu_{X_{k+2}}(y)d\mu_{S'_k}(x)
 =P(\sup\{S'_k,S'_{k+1}\}\leq t),
 \end{align}
    by reversing the above argument. This shows that $\sup\{S_k,S_{k+1}\}$ has the same distribution as $\sup\{S'_k,S'_{k+1}\}$. Using the same arguments above, we can also show $\sup_{n\leq k\leq m}S_k$ has the same distribution as $\sup_{n\leq k\leq m}S'_k$ for all $n\leq m$. We omit the details. 
To show the infinite case, it suffices to note that for $n\in \mathbb N$ and $t\in \mathbb R$,
    \begin{align}
 & P(\sup_{k\geq n}S_k\leq t)\\
 &=P\left(\bigcap_{k=n}^\infty \{S_k\leq t\}\right)\\
 &=\lim_{m\to \infty}P\left(\bigcap_{k=n}^m \{S_k\leq t\}\right)\\
 (\text{by the finite case})&=\lim_{m\to \infty}P\left(\bigcap_{k=n}^m \{S'_k\leq t\}\right)\\
 &=P(\sup_{k\geq n}S'_k\leq t).
 \end{align}
    Hence $\sup_{k\geq n}S_k$ has the same distribution as $\sup_{k\geq n}S_k'$. Thus $\lim_{n\to\infty}\sup_{k\geq n}S_k'=c$ in distribution.
On the other hand, since $\limsup_n S_{n+1}=c$ a.s., we have $\limsup_n S'_n=\limsup_n S_{n+1}-X_1=c-X_1$ a.s., and in particular, in distribution. As weak limits are unique (which can be seen using, say, the L\'evy metric), we have $c=c-X_1$ a.s.
