This question is very closely related to: Continuity in the Krylov-Bogoliubov theorem.
The standard counterexample, which is presented in Katok-Hasselblatt is the following: Let $f:[0,1]\rightarrow [0,1]$ be defined by $f(0)=1$ and $f(x)=x/2$ for $x>0$. Then there is no invariant Borel probability measure.
However, it seems plausible that when a map $f:[0,1] \rightarrow[0,1]$ is piecewise continuous (as the example above) and if the $\omega$-limit set of every point is infinite, then Krylov-Bogoliubov consideration still gives an invariant measure. Is this true?