Here is the definition of an invariant measure that I do know:
Let $(X,\Sigma)$ be a measurable space and let $f\colon X\to X$ be measurable. A measure $\mu$ on $(X,\Sigma)$ is saif to be invariant under f if, for every measurable set $A\in\Sigma$m it is $$ \mu(f^{-1}(A))=\mu(A). $$
Now I have the following definition::
Let $(\Omega,R)$ with $\Omega$ compact metric be a flow. A (regular Borel) measure $\mu$ on $\Omega$ satisfying $\lVert\mu\rVert=1$ is invariant if $$ \int_{\Omega}f(\omega t)\, d\mu(\omega) = \int_{\Omega}f(\omega)\, d\mu(\omega)~~~\text{for all }f\in C(\Omega)~~~\text{ and }t\in R. $$
My problem is that I do not see how this definition of an invariant measure belongs to the definition that I know. Or is there nothing that these both definitions have to do with each other?
By the way: What is meant by $\lVert\mu\rVert=1$?
$\lVert\mu\rVert=1$ probably just means that $\mu$ is a probability measure, i.e. the space $\Omega$ has measure 1.
The two definitions are connected through the fact that the usual definition of $T$-invariant measure (for a discrete action $T:\ X\rightarrow X$) can be rephrased as follows:
$$\int_X f(x)d\mu(x)=\int_X f(T(x))d\mu(x)\qquad \text{for all $f:X\ \rightarrow\mathbb R$ continuous.}$$
This comes from the fact that the condition $\mu(T^{-1}(A))=\mu(A)$ translates to
$$\int_X {1}_A(x) d\mu(x)=\int_A 1\,d\mu(x)=\int_{T^{-1}(A)} 1\, d\mu(x)=\int_X {1}_{T^{-1}(A)}(x)d\mu(x)=\int_X {1}_{A}(T(x))d\mu(x)$$
and that linear combinations of indicator functions ${1}_A$ (with $A$ in the sigma-algebra corresponding to $\mu$) can be used to build up approximations of any continuous function.