Let $A \subset [0,1]$ be meagre. I think it is true (i.e. I have a proof) that the set $T = \{f \in C[0,1]: \mu_L(f(A)) = 0\}$ is dense in $C[0,1]$ ($\mu_L$ is the Lebesgue measure). Is the set $T_n = \{f \in C[0,1]: \mu(f(A)) < \frac{1}{n}\}$ open? Intuitively it should be, but I can't see any way to prove it. I had a couple ideas:
- Prove it is open directly. For any $f \in T_n$, finding $\varepsilon > 0$ such that $B(f, \varepsilon) \subset T_n$, but I couldn't see a way to do that.
- Proving the function $F: C[0,1] \to \mathbb{R}^+$, $F(f) = \mu(f(A))$ is continuous. Again, couldn't see any way to do that.
The end goal is to prove that if $A$ is meagre then $T$ is residual. I do have a fairly convincing proof that $T$ is dense, so I just have to find countable open sets whose intersection is $T$ to complete the proof. If what I am trying to do is impossible or there is an easier way, please let me know.
I realized I was little vague on what $\mu$ is in the definition of $T_n$. Since the end goal is to have $$\bigcap_{n \in \mathbb{N}} T_n =\{f \in C[0,1]: \mu_L(f(A)) = 0\}$$where $\mu_L$ is the Lebesgue measure, we need for $\mu$ to be a function on sets such that $\mu(X) = 0$ implies $\mu_L(X) = 0$ and $\mu(f(A))$ is well-defined. I think this means that $\mu$ is in fact not a measure. Maybe we could use the Jordan outer content for $\mu$?