# CDF of a non-decreasing function of random variable

Let $$X$$ be a random variable. If $$f : \mathbb{R} → \mathbb{R}$$ is right-continuous and nondecreasing. Is the following claim true: $$F_{f(X)}(f(\hat{x}))=F_X(\hat{x})$$

I guess that it is correct since, given that $$f$$ is non-decreasing $$x'>x \Leftrightarrow f(x')>f(x)$$, but I have not found any way to prove it or any proof.

If $$f$$ is not strictly increasing then this is usually not true, though it has the right idea, which is to write the event $$f(X) \leq f(x_0)$$ in the form $$X \leq z(x_0)$$ for some suitable $$z(x_0)$$.
$$z(x_0)=\inf \{ x : f(x) \geq f(x_0) \}.$$
Then $$\{ X \leq z(x_0) \}$$ and $$\{ f(X) \leq f(x_0) \}$$ are just the same event.
For a counterexample to your version, consider $$f(x)=\begin{cases} 0 & x<1/2 \\ 1 & x \geq 1/2 \end{cases}$$, $$X \sim U(0,1)$$ and $$x_0=1/4$$. Then $$F_{f(X)}(f(1/4))=F_{f(X)}(0)=1/2 \neq F_X(1/4)=1/4$$.
As far as I know there is no general term relating $$f(x)$$ to $$g(y)=\inf \{ x : f(x) \geq y \}$$ when $$f$$ is right-continuous and nondecreasing. But when $$f$$ is a CDF, $$g$$ is called the quantile function; cf. "Inverse" of nondecreasing, right-continuous function?