Continuous function is constant $\mu$-almost everywhere, then it has to be constant on the topological support of $\mu$

Question

Let $(X,\tau)$ be a topological space with Borel-$\sigma$-Algebra $\mathcal{B}$ and $\mu$ a measure on $\mathcal{B}$ with topological support $\text{supp}(\mu) := \{x\in X \ |\ x \in U \in \tau \Rightarrow \mu (U) > 0\}$. Let $\varphi : X \rightarrow \Bbb{R}$ be a continuous function with $\mu$-almost everywhere $\varphi = 1$.

The claim is: $\varphi = 1$ on $\text{supp}(\mu)$.

It seems easy, but I don't know where to start. What I know is:

There exists a set $\Omega \in \mathcal{B}$, such that $\mu(\Omega^c) = 0$ and $\varphi(\omega)=1 \; \forall \omega \in \Omega$. If $U$ is a neighborhood of $x \in \text{supp}(\mu)$, then $U\cap \Omega \ne \emptyset$.

Answer 1

Assume that $x\in\text{supp}(\mu)$ with $\phi(x)>1$.

Then - because $\phi$ is continuous - some $U\in\tau$ will exist with $x\in U$ and $\phi(y)>1$ for every $y\in U$.

But $\mu(U)>0$ because $x\in\text{supp}(\mu)$ and a contradiction is found with $\phi=1$ $\mu$-a.e..

This also works if $\phi(x)<1$ and the conclusion $\phi(x)=1$ is justified.

Answer 2

Additionally I found a direct argument:

Let $x\in \text{supp}(\mu)$. Because of continuity we know for all $\varepsilon > 0$ there exists a neighbourhood $U$ of $x$, such that $\varphi(U) \subseteq (\varphi(x)-\varepsilon,\varphi(x)+\varepsilon)$. But it holds $U\cap \Omega \neq \emptyset$ for all those $U$, so we have $1\in (\varphi(x)-\varepsilon,\varphi(x)+\varepsilon)$ for all $\varepsilon >0$.

We conclude $\varphi(x) = 1$.