Conclude almost sure convergence (not only along a subsequence) from a $L^p$-convergence result Let $(E,\mathcal E,\mu)$ be a probaiblity space and $A_n,B$ be linear contractions (operator norm at most $1$) on $L^p(\mu)$ for all $p\in[1,\infty]$. Say we know that $$\left\|(A_n-B)f\right\|_{L^p}\xrightarrow{n\to\infty}0\tag1\;\;\;\text{for all }f\in\mathcal L^p(\mu)\text{ and }p\ge1$$ and $$\left\|(A_n-B)f\right\|_{L^\infty}\xrightarrow{n\to\infty}0\;\;\;\text{for all }f\in U\tag2,$$ where $U$ is a dense subspace of $\mathcal L^1(\mu)$ with $(A_n-B)U\subseteq L^\infty(\mu)$ for all $n\in\mathbb N$.
If $\delta>0$, we know that there is a $g\in U$ with $$\left\|f-g\right\|_{L^1}<\delta\tag3$$ and $$|(A_n-B)g|\le\left\|(A_n-B)g\right\|_{L^\infty}\xrightarrow{n\to\infty}0\;\;\;\text{on }E\setminus N\tag4$$ for some $\mu$-null set $N$.

How can we show that $$\mu\left(\left\{\limsup_{n\to\infty}A_nf-\liminf_{n\to\infty}A_nf>\varepsilon\right\}\right)\le\mu\left(\left\{\sup_{n\in\mathbb N}\left|A_n(f-g)\right|>\frac\varepsilon2\right\}\right)\tag5$$ for all $\varepsilon>0$?

This seems to be an application of a basic inequality, but I can't figure it out. Clearly, by $(4)$, $$\limsup_{n\to\infty}A_ng=\liminf_{n\to\infty}A_ng\;\;\;\text{on }E\setminus N\tag6$$ and it seems like we need to insert this in $(5)$. Moreover, $$\limsup_{n\to\infty}A_nf-\liminf_{n\to\infty}A_nf=\limsup_{n\to\infty}A_nf+\limsup_{n\to\infty}(-A_nf)\tag7.$$
 A: By your $(6)$, we have that 
$$\mu\left(\left\{\limsup_{n\to\infty}A_nf -\liminf_{n\to\infty}A_nf>\varepsilon\right\}\right) = \mu\left(\left\{\limsup_{n\to\infty}A_nf - \limsup_{n \to \infty} A_n g + \liminf_{n \to \infty} A_n g-\liminf_{n\to\infty}A_nf>\varepsilon\right\}\right).$$
Now, on that set, we have the inequalities
\begin{align*}
\varepsilon <& \limsup_{n\to\infty}A_nf - \limsup_{n \to \infty} A_n g + \liminf_{n \to \infty} A_n g-\liminf_{n\to\infty}A_nf
\\ \leq & |\limsup_{n\to\infty}A_nf - \limsup_{n \to \infty} A_n g| + |\liminf_{n \to \infty} A_n f-\liminf_{n\to\infty}A_ng|
\end{align*}
Therefore, it will suffice to show that $$|\limsup_{n\to\infty}A_nf - \limsup_{n \to \infty} A_n g|, |\liminf_{n \to \infty} A_n f-\liminf_{n\to\infty}A_ng| \leq \sup_n |A_n(f-g)|.$$
First let's deal with the $\limsup$ term. Since $A_ng$ is a convergent sequence by assumption, we have that 
$$|\limsup A_n f - \limsup A_n g| = |\limsup A_n(f-g)| \leq \sup_n |A_n(f-g)|$$
where the last equality is trivial from the definition of $\limsup$.
The $\liminf$ term can be treated almost identically.
