Let $\varphi:(X,d,\mu)\to \Bbb R$ be a Lipschitz function, where $\mu$ is a probability measure on the metric space $(X,d)$. The median $m_\varphi$ of $\varphi$ is defined as the real number such that$$\mu\left(\{x\in X: \varphi(x) \geqslant m_\varphi\}\right) \ge \frac{1}{2}, \tag{1}$$and$$\mu\left(\{x\in X: \varphi(x) \leqslant m_\varphi\}\right) \ge \frac{1}{2}. \tag{2}$$Show that the median $m_\varphi$ exists, and is unique.
Say $\varphi$ is $L$-Lipschitz, i.e.,$$|\varphi(x) - \varphi(y)| \le L\, d(x,y),$$ for all $x,y\in X$. I'll try the proof of uniqueness first. Suppose $m_\varphi$ and $m_\varphi'$ are both medians of $\varphi$, i.e., they satisfy inequalities $(1)$ and $(2)$. WLOG, assume $m_\varphi < m_\varphi'$. Then,$$\mu(\{x\in X: \varphi(x) \geqslant m_\varphi\}) = \mu(\{x\in X: \varphi(x) \geqslant m_\varphi'\}) + \mu(\{x\in X: m_\varphi'> \varphi(x) \geqslant m_\varphi\}), \tag{3}$$and$$\mu(\{x\in X: \varphi(x) \leqslant m_\varphi'\}) = \mu(\{x\in X: \varphi(x) \leqslant m_\varphi\}) + \mu(\{x\in X: m_\varphi' \geqslant \varphi(x) > m_\varphi\}). \tag{4}$$
The bounds from $(1)$ and $(2)$ do not yield much unless I am missing something. I am unsure how to get started for proof of existence. I haven't yet used the Lipschitzness of $\varphi$; which might be key to both the proofs. Thanks!