Let $(X,\mu)$ be a $\sigma$-finite measure space. For $f\in L^{1}(X,\mu)$, define a signed measure $\nu_{f}$ on $X$ by $$\nu_{f}(A):=\int_{T^{-1}(A)}f \ d\mu$$ for all measurable $A\subset X$. Then $\nu_{f}$ is finite, so by the Radon-Nikodym theorem, there exists a unique (up to $\mu$-a.e. equivalence) element $P_{T}f\in L^{1}(X,\mu)$ such that $$\nu_{f}(A)=\int_{A}P_{T}f \ d\mu$$ for all measurable $A\subset X$. The operator $P_{T}\colon L^{1}(X,\mu)\to L^{1}(X,\mu)$ is called the Perron-Frobenius operator. One can show that $P_{T}$ is a linear contraction w.r.t. $\|\cdot\|_{1}$.
My question: Suppose that $X=[0,1]$, equipped with the Borel sigma algebra and Lebesgue measure. Let $T\colon X\to X$ be a homeomorphism (or, if needed, something stronger). Suppose that $f\colon[0,1]\to\mathbb{R}$ is continuous. Is it then true that $P_{T}f$ is continuous ($\mu$-a.e.)?
I tried to use density of $C[0,1]$ in $L^{1}[0,1]$, but I'm not sure if that is the way to go. Maybe there is an easier way? Any suggestions are greatly appreciated.