Let $u\in L^1(\mathbb{R}^n)$. Based on that alone, can I say anything nice about the following integral? $$ \int\limits_{\mathbb{R}^n}{\sup\limits_{|y|\le h}{|u(x+y)-u(x)|} \text{ d}x} $$ Ideally, the integral exists for small enough values of $h$ and goes to $0$ as $h\rightarrow 0$.
If I can't say anything nice, what conditions on $u$ will let me make my desired conclusion? For example, $u$ being Lipschitz and compactly supported would work, although I'm looking for much weaker conditions.
In case it's not clear, if you assume only $u\in L^1$ then no, you can't say anything nice.
First: Of course in that case you really want to consider the essential supremum, not the sup: If you let $u$ be the characteristic function of the rationals then $u=0$ almost everywhere although $\sup_{|y|<h}|u(x+y)-u(x)|=1$ for every $x$. That's a totally dumb example that we want to rule out.
So instead, say $m_h(x)$ is the essential supremum of $|u(x+y)-u(x)|$ for $y\in(-h,h)$. That makes more sense; at least if $u_1=u_2$ almost everywhere then $m_h(u_1)=m_h(u_2)$ almmost everywhere.
But it doesn't help; there exists $u\in L^1$ such that $m_h(u)(x)=\infty$ for every $x$ and $h$. For example, let $$v(x)=x^{-1/2}\chi_{(0,1)}(x),$$say $(x_n)$ is a countable dense subset of $\Bbb R$, and let $$u(x)=\sum_{n=1}^\infty 2^{-n}v(x-x_n).$$