Vertical spacing suddenly changes as I am writing a document

Question

This has happened a few times and I don't know how to diagnose the issue. Below is the code which is as simple as I can get to reproduce the problem.

The vertical spaces between a lot of the lines is far too large suddenly. Removing something like the title or the description environment suddenly fixes it. I cannot target the problem.

\documentclass{amsart}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}

\newenvironment{problem}[2][Problem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}

\title{title}
\author{author}

\begin{document}
\maketitle

\begin{problem}{2.32}
    Suppose that $\mu(X)<\infty$. If $f$ and $g$ are complex-valued measurable functions on $X$, define:
        \[\rho(f,g) := \int\frac{|f-g|}{1+|f-g|}\,d\mu.\]
    Show that $\rho$ is a metric on the space of measurable functions if we identify functions which are equal $\mu$-a.e.. Moreover, $f_n\to f$ with respect to $\rho$ if and only if $f_n\to f$ in measure.
\end{problem}
\begin{proof}
    First we need to show that $\rho(f,g)<\infty$ and that if $f$ and $f'$ are equal $\mu$-a.e., then $\rho(f,g) = \rho(f',g)$. Note that the integrand of $\rho$ is bounded above by 1 and $X$ is of finite measure, thus $\rho(f,g)$ is finite for any $f$ and $g$ (being bounded above by the integral of $\chi_X$). Then note:
    \[|f-g| = |f-g| + |f'-f| \geqq |f-g+f'-f| = |f'-g|, \text{ $\mu$-a.e.},\]
    and similarly, we can get the opposite inequality by swapping roles of $f$ and $f'$. Thus $|f-g| = |f'-g|$, $\mu$-a.e. and then the well-definedness of $\rho$ follows as the integrands of $\rho(f,g)$ and $\rho(f',g)$ are $\mu$-a.e. equal.

Now to prove that it is a metric:
    \begin{description}
        \item[Indiscernibility] Note that if $f=g$ $\mu$-a.e., then clearly $|f-g| = 0$ $\mu$-a.e., and hence $\rho(f,g)$ is the integral of a $\mu$-a.e. zero function---hence it is zero. On the other hand, if $\rho(f,g) = 0$, then we know that the integrand (being in $L^+$) is zero, $\mu$-a.e.. Thus $|f-g|$ is zero, $\mu$-a.e. and thus $f=g$ $\mu$-a.e.. Hence:
            \[\rho(f,g) = 0 \iff f = g, \mu\text{-a.e.}.\]
        \item[Symmetry] Note that $|\cdot-\cdot|$ is symmetric, and thus $\rho$ inherits this symmetry as the integrand is symmetric in $f$ and $g$. It also might be worth mentioning that the integrand is positive, so $\rho\geqq 0$.
        \item[$\blacktriangle$-inequality] Note\footnote{Meanwhile also noting that $0\leqq x<y\implies \frac{x}{1+x} < \frac{y}{1+y}$. To deduce this, just take its derivative to note it is increasing.}:
            \begin{align*}
            \rho(f,h)
            &= \int\frac{|f-h|}{1+|f-h|}\,d\mu\\
            &= \int\frac{|f-g+g-h|}{1 + |f-g+g-h|}\,d\mu\\
            &\leqq \int\frac{|f-g| + |g-h|}{1 + |f-g| + |g-h|}\,d\mu\\
            &= \int\frac{|f-g|}{1 + |f-g| + |g-h|}\,d\mu + \int\frac{|g-h|}{1+ |f-g| + |g-h|}\,d\mu\\
            &\leqq \int\frac{|f-g|}{1 + |f-g|}\,d\mu + \int\frac{|g-h|}{1+ |f-g|}\,d\mu\\
            &= \rho(f,g) + \rho(g,h).
            \end{align*}
    \end{description}
\end{proof}
\end{document}

A comparison of before and after commenting out the \maketitle:

Maybe it has something to do with page breaks...?