Let $E\subset \Bbb R^d$ be measurable, and let $\chi_E$ be the characteristic function: $$\chi_E(x)=\begin{cases}1,&x\in E\\0,&x\not\in E\end{cases}$$
I want to show that for $f_h(x)=x-h$ that $\chi_E\circ f_h\to \chi_E$ as $h\to 0$ in the $L^1(\Bbb R^d)$-norm.
My approach so far is: $$|\chi_E\circ f_h - \chi_E|=|\chi_{E-h} - \chi_E| = \begin{cases}1,& x\in (E-h),x\not\in E\\ 0,& x\in (E-H)\cap E\\1,& x\in E,x\not\in E-h\end{cases}$$ $$=\chi_{(E-h)\Delta E}$$ where $A\Delta B$ is the symmetric difference $A\Delta B = A\cap B^c \cup A^c \cap B$.
Thus: $$\|\chi_E\circ f_h-\chi_E\|=\int \chi_{(E-h)\Delta E)} = m((E-h)\Delta E)$$ and it's 'clear' that $(E-h)\Delta E= \emptyset$ for $h=0$, so I feel like I can conclude $$\lim_{h\to 0}\|\chi_E\circ f_h - \chi_E\|=\lim_{h\to 0} m((E-h)\Delta E)=m(\emptyset)=0,$$ but this doesn't seem rigorous, and I can't seem to actually write down what $m((E-h)\Delta E)$ is.
How does one properly conclude that $\chi_E\circ f_h\to \chi_E$ as $h\to 0$?