Explain why $\liminf\frac{1}{n}\sum\limits^n_{i=1} X_i=\liminf\frac{1}{n}\sum\limits^n_{i=N} X_i$

Question

Let $(X_n)_{n \in \mathbb{N}}$ be real random variables.

Then, why do we have

$$\liminf_{n\rightarrow \infty} \frac{1}{n}\sum^n_{i=1} X_i=\liminf_{n\rightarrow \infty} \frac{1}{n}\sum^n_{i=N} X_i$$ for any $N\in \mathbb{N}$ ?

Any help would be appreciated.

Edit: The thing is that I'm reading this as $\liminf_{n\rightarrow \infty} \frac{1}{n}\sum^n_{i=1} X_i=\sup_{n\geq 1} \inf_{m\geq n} \frac{1}{m}\sum^m_{i=1} X_i$

and I don't see how $$\sup_{n\geq 1} \inf_{m\geq n} \frac{1}{m}\sum^m_{i=1} X_i=\sup_{n\geq 1} \inf_{m\geq n} \frac{1}{m}\sum^m_{i=N} X_i$$. Or maybe I'm just 'translating' wrongly the expression.

Answer 1

(2017/8/6) "Amusing" silent downvote, surely for mathematical reasons... :-)

Because, if $z_n\to\ell$ then $\liminf(y_n+z_n)=(\liminf y_n)+\ell$. Your question is concerned with the case $y_n=\frac1n\sum\limits_{k=N}^nX_k(\omega)$ and $z_n=\frac1n\sum\limits_{k=1}^{N-1}X_k(\omega)$, hence $\ell=0$.