I'm reading a paper by Micheal Handel, Bruce Kitchens and Daniel J. Rudolph on Entropy <http://link.springer.com/article/10.1007/BF02761650> I have some problem 


I couldnt show that $\mu(\bigcup\_{N\ge1}G\_N(C))=1 $ in following lemma


**Lemma:**


If $(X,d)$ be non compact metric space and $\mu$ be Borel probability measure on $X$ and invariant under $T:X\to X$ and compact set $C$ with $\mu(C) > 1-\delta/2$


There is a compact set $K \subseteq C$ with $\mu(K) > 1 - \delta$ and an $N$ so that for every
$n\ge N$ and $x \in K$
$$\frac{1}{n}\sum\_{i=0}^{n-1}\mathbb{1}\_{C}(T^{i}x)\ge 1-\delta$$
where $\mathbb{1}\_{C}$ is characteristic function.


to prove this lemma author is used following process.


For each $N\ge1$ we define


$$G\_N(C)=\{x \in X :\frac{1}{n}\sum\_{i=0}^{n-1}\mathbb{1}\_{C}(T^{i}x)\ge 1-\delta,
\forall n\ge N \}$$


and take $K=C\cap G\_N(C)$


**1.** $G\_N(C)$ is closed so $K$ is compact 


but I couldn't show that $\mu(K) > 1 - \delta$



Question about mean sojourn time on non compact space
**[Edited]** after comment by the asker and corrected a computation.


I think you can use the [ergodic theorem](https://en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorems) for merely invariant measures, *but the lemma might need some adaptation* (without access to the article, I cannot tell whether such adaptation hurts or not -- you should link to openly accessible version of the article whenever available).




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**[Further edit]** Indeed here is a *counter-example to the original lemma.* Let $T:\mathbb{S}^1\times [0,1]\to \mathbb{S}^1\times [0,1]$ be defined by $T(x,y)=(x+\theta,y)$ where $\theta$ is an irrational number, and let $\mu$ be the Lebesgue measure. For any small $\delta$, let $C$ be the union of $[0,1-\frac{11}{10}\delta]\times [0,1/2]$ and $\mathbb{S}^1\times [1/2,1]$; $C$ has measure $1/2-\frac{11}{20}\delta+1/2>1-\delta/2$. On the other hand, the set $G\_N(C)$ consist of only $\mathbb{S}^1\times [1/2,1]$ so that you cannot get $K$ with measure more than $1/2$, however small was $\delta$ in the first place.




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Here is a proof of a weakened Lemma, which might be sufficient to get the main conclusion from the article.


Let us assume that $\mu(C)>1-\alpha\delta$, and then tailor the $\alpha$ to get the desired conclusion.


The time averages converge in $L^1$ to some invariant measurable function $g$, which might not be constant since $\mu$ is not assumed to be ergodic, but which is positive and less than 1 by construction and has the same integral than the original function by the ergodic theorem. This means it has integral at least $1-\alpha\delta$, and since it cannot exceed one it must be close to $1$ on a large set. More precisely, 
$$1-\alpha\delta<\int g d\mu \le \mu(\{g\ge1-\delta\})+(1-\delta)(1-\mu(\{g\ge1-\delta\})) \\=1-\delta+\delta \mu(\{g\ge1-\delta\})$$
so that $\mu(\{g\ge1-\delta\})> 1-\alpha$ (my originial computation had a sign error, I assumed the signed that was convenient was the correct one - have to learn from that).


Note that you don't get $\mu(\cup G\_N(C))=1$, and actually **[edited]** you *cannot* expect it: if $T:\{0,1\}\to\{0,1\}$ is the identity map and $C=\{0\}$ then $G\_N(C)=C$ for all $N$. 


Now for $N$ large enough the finite time-average must be larger than $1-\delta$ on a set of size $1-\alpha$, i.e. $\mu(G\_N(C))> 1-\alpha$ (apply the above displayed computation to $x\mapsto \frac1N\sum\_{0}^{N-1} \mathbb{1}\_C(T^ix)$ instead of $g$, with $N$ large enough that the integral is bounded below by $1-\alpha\delta$). Taking $K=C\cap G\_N(C)$ you get $\mu(K)\ge 1- \alpha\delta-\alpha$. To have $\mu(K)\ge 1-\delta$, we take $\alpha=\delta/(1+\delta)$ i.e. assume $\mu(C)>1-\delta^2/(1+\delta)$, which is *stronger* than the original assumption.


