Enciclopedia of mathematics says the following:
"A is called a null (or negligible) set if $\mu_∗(A)=0$; in this case the complement $X\setminus A$ is called a set of full measure (or conegligible), and one says that $x\notin A$ for almost all $x$ (in other words, almost everywhere). Two sets $A,B\subset X$ are almost equal (or equal mod $0$) if $(x\in A) \Leftrightarrow (x\in B)$ for almost all $x$ (in other words, $A\setminus B$ and $B∖A$ are negligible). Two functions $f,g:X \longrightarrow Y$ are almost equal (or equal mod 0, or equivalent) if they are equal almost everywhere. "
https://www.encyclopediaofmath.org/index.php/Measure_space
Ok guys i think proved it, i used the following preposition which is equivalent for a dynamical topological pair $(X,T)$ to be uniquely ergodic :
For every continuous function $f\in C(X)$
$$\frac{1}{n}\sum_{k=0}^{n-1}f(T^k(x)) \to \int fd\mu \tag 1$$
for every $x\in X$ where $\mu$ is the uniquely determined $T-$invariant Borel probability measure.
Now i can prove the the desired result as follows:
Suppose we have a set $F$ which closed, $T-$invariant $\bigl(T(F)\subseteq F\bigr)$ and $F \neq X$ then $X\setminus F$ is non empty and open. Let $x_1 \in X\setminus F$, then there is an $\epsilon>0$ such that
$$E=\hat{B}(x_1,\epsilon) = \{x\in X: d(x,x_1)\leq \epsilon\} \subseteq X\setminus F$$
From our assumptions we get that $\mu(E)>0 $ since $\hat{B}(x_1,\epsilon)$ contains the open ball $B(x_1,\epsilon).$
Now we are trying to find the crucial continuous function $f$ in order to use $(1)$.Since we have two closed and disjoint sets $F,E$ we define
$$f(x) = \frac{dist(x,F)}{dist(x,F)+dist(x,E)}$$
then we know that
$\alpha)$ $f$ is continuous
$\beta)$ $0\leq f \leq 1$
$\gamma$) $f|F =0$ and $f|E=1$
So if we combine these facts and $(1)$ we get that
$$\frac{1}{n}\sum_{k=0}^{n-1}f\bigl(T^k(x)\bigr) \to \int fd\mu=\int_{X\setminus F}f d\mu \geq \int_{E}fd\mu \geq \mu(E)>0 \tag 2$$
for every $x\in X$.
But now (2) tells us that $F$ must be the empty set since if we can find $x^*\in F$ since $T(F) \subseteq F$ we get that $T^k(x^*) \in F$ for every k. In other words
$$\frac{1}{n}\sum_{k=0}^{n-1}f\bigl(T^k(x^*)\bigr) =0 $$
for every $n \in \mathbb{N}$ which contradicts $(2)$ since it holds for every $x\in X$.
So, for every $F$ closed which is $T-$invariant $\bigl(T(F)\subseteq F\bigr)$ it must be either $F= \emptyset$ either $F=X$.So, $(X,T)$ must be minimal.
Is this correct? thanks again !
Best Answer
If the space has a countable basis and for the measure $\mu$ any open set has positive measure we can have this result as proved below, in other case there may be counterexample.
Since $X$ is a metric space, we can find countable basis $\{U_j\}_{j=1}^{\infty}$. Not that if for $x\in X$, the orbit of $x$ is not dense, then there exists some $U_j$ such that $\{T^n(x):n\in \mathbb{Z}\}\cap U_j=\emptyset$ so $$x\in A_j=X-\cup_{i=0}^{\infty}T^{-i}U_j.$$ So the set of $X$ such that the orbit of $x$ is not dense is contained in the set $\cup_{j=1}^{\infty} A_j$. Note that $A_j$ is $T$-invariant and has a empty intersection with $U_j$ so $\mu(A_j)=0$ since $U_j$ has positive measure. And so $\mu(\cup_{j=1}^{\infty} A_j)=0$ and the proof is completed.