I wish to prove the following
If $f$ is integrable and $f:\mathbb{R}\to\mathbb{R}$ then $$\lim_{t\to0}\int_{-\infty}^{\infty}|f(x)-f(x+t)| = 0$$ I have in my notes that if there exists a $g(x) \in \mathbb{L}(\mathbb{R})$ such that for every $t \in [a,b]$ $$ |f(x,t)| \leq g(x) $$ then $$\lim_{t\to t_0}\int f(x,t) = \int \lim_{t \to t_0}f(x,t)$$ We proved the above result in class.
So my first question is, what can I choose for a g(x) that satisfies the condition for switching the limit inside and outside the integral?
And assuming that I can find a $g(x)$ can someone please check the next step(s).
$$\lim_{t\to0}\int_{-\infty}^{\infty}|f(x)-f(x+t)| = \int_{-\infty}^{\infty}\lim_{t\to0}|f(x)-f(x+t)|$$
Let $t_n$ be any sequence in $\mathbb{R}$ such that $t_n \to 0$, then the above is equivalent to saying $$\int\lim_{n\to\infty}|f(x)-f(x+t_n)| = \int |f(x)-f(x+0)|$$ $$= \int0 = 0$$
Is my general idea of how to solve the problem correct even if I cant find a $g(x)$? What would $g(x)$ be?
Edited: Here is my attempt with Jake's suggestion -
I'm trying it for a characteristic function only to make sure I have the right idea.
Let $f = \chi_A$ with $m(A)<\infty$
and let $g = f(x+t) = \chi_{A-t}$
Define $h:= |f-g|$ as $\chi_C$ where $$C = \{x\notin A\}\cap\{x-t \in A\}\bigcup\{x\in A\}\cap\{x-t \notin A\}$$ and let $t_n$ be any sequence in $\mathbb{R}$ such that $t_n \to 0$. As $t_n \to 0, \{x\in A\}\cap\{x-t_n \notin A\}$ becomes the empty set and so does $\{x\notin A\}\cap\{x-t \in A\}$.
Thus $\chi_C \equiv 0$ on $\mathbb{R}$ and $$\int h = \int0 = 0$$
Is this right?