Is the median preserved for any strictly monotonic mapping? Problem
Let $X\sim f_X$, where $f_X$ is the probability density function of $X$. Let $g: \mathbb{R} \to \mathbb{R}$ a strictly monotonic (decreasing or increasing) mapping. I aim to prove or disprove:
$$
g[\mathrm{median} (X)] = \mathrm{median}[g(X)]
$$

Try
Let $x_0 = \mathrm{median} (X)$, $Y=g(X)$, $y_0 = g(x_0)$. Then, we want to show:
$$
0.5 = \int_{-\infty}^{y_0} f_Y (y) dy
$$
WLOG, assume $g$: increasing. By change of variables, $f_Y(y) = f_X(x)/g'(x)$ and $dy=dx g'(x)$. Thus,
$$
\begin{align}
\int_{-\infty}^{y_0} f_Y (y) dy &= \int_{-\infty}^{y_0} f_X(x) dx \\
&\neq \int_{-\infty}^{x_0} f_X(x) dx = 0.5
\end{align}
$$
So, if $f_X>0$ a.s., then the above statement holds only if $g(x_0) = x_0$. However, take an example of log-normal distribution, where $X\sim N(\mu, \sigma^2)$ and $g(x)=\exp(x)$. There is no real solution to $g(x_0) = x_0$, and $f_X >0$ a.s. But it is still true that $\mathrm{median}[g(X)] = g(\mu)$.
So my disproof is invalid? Any suggestions are welcome.

By definition, $m$ is a median of the random variable $X$ if and only if $$\mathrm{Pr}[X ≤ m] ≥ \tfrac12 \text{ and } \mathrm{Pr}[X ≥ m] ≥ \tfrac12.$$
(Note that, in general, $X$ may have multiple medians if there is an interval of values satisfying the definition, and that it's possible for either or both probabilities to be strictly greater than $\tfrac12$ if $\mathrm{Pr}[X = m] > 0$.)
Also by definition, a function $g: \mathbb R \to \mathbb R$ is monotone increasing if and only if $$a ≤ b \implies g(a) ≤ g(b)$$ for all $a,b \in \mathbb R$.
(A function $g$ is called strictly monotone increasing if $a < b \implies g(a) < g(b)$, but we don't actually need this stronger property here.)

Now, let $m$ be a median of $X$ and let $g$ be monotone increasing.  Then $X ≤ m \implies g(X) ≤ g(m)$ (and $X ≥ m \implies g(X) ≥ g(m)$), and thus $$\begin{aligned}
\mathrm{Pr}[g(X) ≤ g(m)] ≥ \mathrm{Pr}[X ≤ m] &≥ \tfrac12 \text{ and } \\
\mathrm{Pr}[g(X) ≥ g(m)] ≥ \mathrm{Pr}[X ≥ m] &≥ \tfrac12.
\end{aligned}$$
In other words, if $m$ is a median of $X$ and $g$ is monotone increasing, then $g(m)$ is a median of $g(X)$.

Ps. Note that the result above holds for any real-valued random variable $X$ and any monotone increasing function $g$, even if the medians of $X$ and/or $g(X)$ are not uniquely defined.
However, it only guarantees that $g(m)$ is some median of $g(X)$, but not that any particular rule for picking a "canonical" median $m$ for $X$ — such always taking the low end, high end or the midpoint of the interval of possible medians — will necessarily yield $g(m)$ when applied to the distribution of $g(X)$.
Also, if the function $g$ is discontinuous, $g(X)$ may have multiple medians even if $X$ does not, while if $g$ is not strictly increasing, $X$ may have multiple medians even if $g(X)$ does not.  And if $g$ is not strictly increasing, then it's also possible for $g(m)$ to be a median of $g(X)$ even if $m$ is not a median of $X$.
In particular, even if we define $\operatorname{median}[X]$ to be the set of all medians of $X$, the most we can say in general is that $g(\operatorname{median}[X]) \subseteq \operatorname{median}[g(X)]$.

Pps. The result above can also be straightforwardly generalized to arbitrary quantiles by defining $m$ to be a $p$-quantile of $X$ (for some $p \in [0,1]$) if and only if $$\mathrm{Pr}[X ≤ m] ≥ p \text{ and } \mathrm{Pr}[X ≥ m] ≥ 1-p$$ and using essentially the same proof to show that, if $g$ is monotone increasing, then $g(m)$ is a $p$-quantile of $g(X)$.
