I'm trying to costruct a sequence $(g_{n})_{n}$ such that it converges a.e to zero and $$\limsup\int g_{n}d\lambda=1,$$ $$\liminf\int g_{n}d\lambda=-1$$ with respect to Lebesgue measure.
I think $(g_{n})_{n}$ has to be a three step fuction of the form
$$g(x)_{n}=\left\{\begin{matrix} -1 & \\ 0& \\ 1& \end{matrix}\right.$$
but I'm not quite sure what the support of each step for $g(x)_{n}$ has to be.
Any idea would be helpful.
On
One strategy here is to start with a nice example of a sequence of functions that converge pointwise or almost everywhere to zero, but which fail to integrate to zero in the limit. For example, consider the sequence $$ f_n = n \chi_{[0,1/n]}. $$
Observe that $$ \int f_n = \int n \chi_{[0,1/n]} = n \mu([0,\tfrac{1}{n}]) = 1 $$ for all $n \in \mathbb{N}$. On the other hand, for all $x\ne 0$, we have $f_n(x) \to 0$. Thus $f_n \to 0$ almost everywhere, but $\lim_{n\to\infty} \int f_n = 1$. Now all we need to do is adjust the sign. The usual trick is to multiply by $-1$ for odd $n$, i.e. define $$ g_n = (-1)^n f_n = (-1)^n n \chi_{[0,1/n]}. $$ It can be verified that $$ \int g_n = \begin{cases} 1 & \text{if $n$ is even, and} \\ -1 & \text{if $n$ is odd}, \end{cases} $$ from which the desired result follows.
Consider $$ g_n= \begin{cases}I_{[n,n+1]}&\text{$n$} \,\text{even}\\ -I_{[-n-1,-n]} &n \,{\text{odd}} \end{cases} $$ for $n\geq1$.