Composition of a Continuous and Measurable Function

[Bashyboy]

Homework Statement

Suppose that $f$ and $g$ are real-valued functions defined on all of $\Bbb{R}$,$f$ is measurable, and $g$ is continuous. Is the composition $f \circ g$ necessarily measurable?

Homework Equations

The Attempt at a Solution

Let $c \in \Bbb{R}$ be arbitrary. Then $\{x \in \Bbb{R} ~|~ f(g(x)) > c \} = \{x \in g( \Bbb{R}) ~|~ f(x) > c \} \subseteq \{x \in \Bbb{R} ~|~ f(x) > c\}$, where the last set is measurable by our hypothesis. If $g$ were surjective, the first set would certainly be measurable, independently of whether $g$ is continuous, since equality would hold with the last set. This suggests that the statement isn't necessarily true. Another way would be to argue that $g(\Bbb{R})$ is a measurable, but apparently the continuous image of a measurable set isn't always measurable. So I suspect the statement isn't necessarily true, but I am having trouble finding a counterexample. I could use some help.

 

[Bashyboy]

Hold on! I am pretty certain $\{x \in \Bbb{R} ~|~ f(g(x)) > c \} = \{x \in g( \Bbb{R}) ~|~ f(x) > c \}$ is false. However, I still believe that the theorem isn't necessarily true.

 

[fresh_42]

I'm not really fit in the dungeon of topological, pathological examples, but I've seen an example which showed that the composition of two Lebesgue measurable functions isn't necessarily Lebesgue measurable. The author constructs a continuous, $1:1$, strictly increasing function $g$ on $[0,1]$ and with the help of a nowhere dense perfect subset $P$, $f=\chi_A$ where $A=g(S)$ and $S \subseteq P$ is a non Lebesgue measurable subset, $f \circ g$ does the job.