About Birkhoff averages I have the following problem
Let $(X,\mathcal{A},\mu, T)$ be a measure-preserving dynamical system, and let $f\in L^1(\mu)$. Let $\tilde{f}(x)$ be the limit of the Birkhoff averages of $f$ (defined almost everywhere). Show that $$\tilde{f}(x)=\lim_{\lambda\to 1^-}(1-\lambda)\sum_{j=0}^{\infty}\lambda^jf(T^j(x))\quad for\thinspace a.e.\thinspace x.$$
Honestly I don't know how to proceed.
If anyone can give a Hint it will be appreciated !
Thanks !
 A: Let me add to Dap's hint and Michael's comments that the dynamical system is a distraction in this problem.  The simpler and more general statement is to that Cesàro summability implies Abel summability (with the same value).
Namely, let $a_0,a_1,\ldots$ be a sequence of real numbers and assume that the Cesàro averages $S_n:=\frac{1}{n}(a_0+a_1+\cdots+a_{n-1})$ converge to a number $\overline{a}$ as $n\to\infty$.  The claim is that the power series $(1-\lambda)\sum_{j=0}^\infty\lambda^j a_j$ converges to $\overline{a}$ as $\lambda\uparrow 1$.
This can be proven using Dap's hint.  Now from the ergodic theorem, the sequence $f(x), f(T(x)), f(T^2(x)), \ldots$ is almost everywhere Cesàro summable to a function $\tilde{f}(x)$.  Therefore, it is also Abel summable to the same value $\tilde{f}(x)$.
A: Here are some details on importance of the measure preserving assumption that Dap added as an edit for the question. 
The special case of measure-preserving $T$
In general, $f \in L^1(\mu)$ does not imply $f(T) \in L^1(\mu)$. However, it is true in the special case when $T$ is measure-preserving (so that $\mu(A) = \mu(T^{-1}(A))$ for all measurable sets $A$). If $T$ is measure-preserving  and if $h$ is any measurable and integrable function, it can be shown that: 
$$ \int h(x)d\mu = \int h(T^j(x)) d\mu \quad, \forall j \in \{0, 1, 2, \ldots\} $$
Intuition on weighted averages:
If $f \in L^1(\mu)$ and $T$ is measure-preserving, then there is an $a \in \mathbb{R}$ such that: 
$$ \int f(T^j(x)) d\mu = a \quad, \forall j \in \{0, 1, 2, \ldots\} $$
It follows that if $\{w_j\}_{j=0}^{\infty}$ are any nonnegative weights such that $\sum_{j=0}^{\infty} w_j = 1$, then 
$$ \int \left[\sum_{j=0}^{\infty} w_j f(T^j(x)) \right]d\mu = \sum_{j=0}^{\infty} w_j a  = a  $$
where passing the integral through the infinite sum can be formally justified by the Lebesgue dominated convergence theorem using the integrable bounding function $\sum_{j=0}^{\infty} w_j |f(T^j(x))|$. 
This holds for any type of weighted average, including a sum like $\frac{1}{n}\sum_{j=0}^{n-1}f(T^j(x))$ or an exponentially weighted average like $(1-\lambda)\sum_{j=0}^{\infty} \lambda^j f(T^j(x))$. 
Further, since the integral is finite, it follows that $\sum_{j=0}^{\infty} w_j f(T^j(x))$ is well defined and finite for almost all $x$ (except possibly a set of measure zero). 
More on the Dap hints
As  Dap mentions, the measure-preserving assumption ensures an ergodic-type  theorem that, for almost all $x$, ensures the sequence $\{m_n\}$ converges to some real number $m$, where both $m_n$ and $m$ can depend on $x$ and where $m_n$ is defined for all $n \in \{1, 2, 3, ...\}$ by 
$$ m_n = \frac{1}{n}\sum_{j=0}^{n-1} f(T^j(x)) $$
Dap’s hint of writing $(1-\lambda) \sum_{j=0}^{\infty} \lambda^j f(T^j(x))$ in terms of the $m_n$ values is a good one and you should follow that through.  However, the resulting argument is still non-trivial. It will help to remember that since the values $m_n$ converge, those values are also bounded.  Also, it may help to eventually rewrite the summation in a different order and/or  break it up, if needed, into terms that have one behavior and terms that have another.  
