Show that $\liminf \limits _{k\rightarrow \infty} f_k = \lim \limits _{k\rightarrow \infty} f_k$ Is there a way to show that $\liminf \limits _{k\rightarrow \infty} f_k = \lim \limits _{k\rightarrow \infty} f_k$. The only way I can think of is by showing $\liminf \limits _{k\rightarrow \infty} f_k = \limsup \limits _{k\rightarrow \infty} f_k$. Is there another way?
Edit: Sorry, I should have mentioned that you can assume that $\{f_k\}_{k=0}^\infty$ where $f_k:E \rightarrow R_e$ is a (Lebesgue) measurable function
 A: What you have is incorrect. For instance, consider $$f_k = \begin{cases} 0 & \text{if }k \text{ is even}\\1 &\text{if }k \text{ is odd}  \end{cases}$$
$\liminf f_k = 0$, $\limsup f_k = 1$ while $\lim f_k$ does not exist.
A: The following always holds: $\inf_{k\geq n} f_k \leq f_n \leq \sup_{k\geq n} f_k$. Note that the lower bound in non-decreasing and the upper bound is non-increasing.
Suppose $\alpha = \liminf_k f_k = \limsup_k f_k$, and let $\epsilon>0$. Then there exists a $N$ such that for $n>N$, we have $\alpha -\inf_{k\geq n} f_k < \epsilon$ and $\sup_{k\geq n} f_k -\alpha < \epsilon$. Combining this with the above inequality yields $-\epsilon < f_k - \alpha< \epsilon$ from which it follows that $\lim_k f_k = \alpha$.
Now suppose $\alpha = \lim_k f_k$. Let $\epsilon >0$, then there exists a $N$ such that $-\frac{\epsilon}{2}+\alpha < f_k< \frac{\epsilon}{2}+\alpha$. It follows from this that $-\epsilon + \alpha \leq \inf_{k\geq n} f_k \leq \sup_{k\geq n} f_k < \epsilon+\alpha$, and hence $\liminf_k f_k = \limsup_k f_k = \alpha$.
Hence the limit exists iff the $\liminf$ and $\limsup$ are equal.
