I asked myself the following question, which I think should be true but I cannot come up with a good reasoning why it does or doesn't hold.
Let $K$ be a compact operator on $L^p(\mathbb{R})$. Do we get then $$ M_{\mathbb{1}_{[-A,A]}} K \to K \quad \text{or} \quad KM_{1_{[A,A]}} \to K \quad \text{for } A \to \infty,$$ where the convergence is meant in the operator norm sense and $M_{1_{[A,A]}}$ denotes the operator of multiplication by the characteristic function of the interval $[-A,A]$?
I'll consider $B_A = I - M_{\mathbf{1}_{[-A,A]}}$, then the questions are whether $B_AK \to 0$ and $KB_A \to 0$ respectively.
Consider $B_AK$ first. If $1 \leqslant p < \infty$, then $B_A$ converges to $0$ pointwise ($B_A f \to 0$ for all $f\in L^p(\mathbb{R})$). Since the operators $\{ B_A : A \in (0,+\infty)\}$ are equicontinuous - $\lVert B_A\rVert = 1$ for all $A$ - it follows that the convergence is uniform on compact sets. By definition of a compact operator, $\overline{K(U)}$ is compact, where $U$ is the unit ball of $L^p(\mathbb{R})$, so $B_A$ converges to $0$ uniformly on $K(U)$, and that is
$$0 = \lim_{A\to \infty} \sup\: \bigl\{ \lVert B_Ag\rVert : g \in K(U)\bigr\} = \lim_{A \to \infty} \sup\: \bigl\{ \lVert B_AKf\rVert : f \in U\bigr\} = \lim_{A \to \infty} \lVert B_AK\rVert.$$
For $p = \infty$, we don't have $B_Af \to 0$ for all $f\in L^{\infty}(\mathbb{R})$ and thus no good reason to expect that $B_AK \to 0$. Indeed, let $\lambda \in L^{\infty}(\mathbb{R})$ a nonzero continuous linear functional, and let $Kf = \lambda(f)\cdot \mathbf{1}_{\mathbb{R}}$. Since $K$ is continuous and has finite rank, $K$ is compact, but $\lVert B_AK\rVert = \lVert K\rVert = \lVert\lambda\rVert$ for all $A$.
Now consider $KB_A$. Suppose
$$\limsup_{A \to \infty} \lVert KB_A\rVert > \varepsilon > 0.$$
Then there is an increasing sequence $A_n \to \infty$ and a sequence $(f_n)$ in the closed unit ball of $L^p(\mathbb{R})$ with
$$\lVert KB_{A_n} f_n\rVert \geqslant \varepsilon.$$
Set $g_n = B_{A_n}f_n$. By compactness of $K$, we may assume that $Kg_n \to h$ in $L^p(\mathbb{R})$, and it follows that $\lVert h\rVert \geqslant \varepsilon$.
But if $1 < p \leqslant \infty$, then $g_n \to 0$ weakly, and hence $Kg_n \to 0$ weakly. Thus $\lVert KB_A\rVert \to 0$ for $1 < p \leqslant \infty$. And if $p = 1$, let
$$K(f) = \Biggl(\int_{\mathbb{R}} f\,d\lambda\Biggr)\cdot \varphi,$$
where $\varphi$ is your favourite nonzero function in $L^1(\mathbb{R})$. Then
$$\lVert KB_A\rVert = \lVert K\rVert = \lVert\varphi\rVert$$
for all $A$, so $\lVert KB_A\rVert \not\to 0$ in this case.
We assume $1\leq p<\infty$.
Let $f\in L^p(\mathbb R)$. Fix $\varepsilon>0$. Then there exists $A>0$ such that $\displaystyle\int_{|x|>A}|f|^p<\varepsilon.$ So, if $I$ denotes the identity operator, $$ \|(I-M_{1_{[-A,A]}})f\|_p^p<\varepsilon. $$ That is $M_{1_{[-A,A]}}\to I$ pointwise. It follows that $$ \|M_{1_{[-A,A]}}F- F\|\to0 $$ for all finite-rank operators $F$. Also $\|M_{1_{[-A,A]}}\|=1$.
Now given a compact $K$, and $\varepsilon>0$, there exists $F$ finite-rank with $\|F-K\|<\varepsilon$. Then \begin{align} \|M_{1_{[-A,A]}}K- K\|&\leq\|M_{1_{[-A,A]}}(K- F)\|+\|M_{1_{[-A,A]}}F- F\|+\|F-K\|\\ \ \\ &\leq2\varepsilon+\|M_{1_{[-A,A]}}F- F\| \end{align} Thus $\limsup_A \|M_{1_{[-A,A]}}K- K\|=0$, and so $\|M_{1_{[-A,A]}}K- K\|\to0$.
For finite-rank $F$, pointwise convergence implies norm convergence. Then, as $$ \|FM_{1_{[-A,A]}}f-Ff\|\leq\|F\|\,\|M_{1_{[-A,A]}}f-f\|\to0, $$ we get that $\|FM_{1_{[-A,A]}}-F\|\to0$. Given $K$ compact and $\varepsilon>0$, we take $F$ finite-rank with $\|F-K\|<\varepsilon$. Then \begin{align} \|KM_{1_{[-A,A]}}-K\|&\leq\|KM_{1_{[-A,A]}}-FM_{1_{[-A,A]}}\| +\|FM_{1_{[-A,A]}}-F\|+\|F-K\|\\ \ \\ &\leq2\varepsilon+\|FM_{1_{[-A,A]}}-F\|, \end{align} and again we deduce that $\|KM_{1_{[-A,A]}}-K\|\to0$.
The limit does not hold when $p=\infty$. For instance, choose some bounded linear functional $\phi$ on $L^\infty(\mathbb R)$ with $\phi(1)=1$, and define $Kf=\phi(f)\,1$. Then $K$ is finite-rank, so compact, and $$ \|M_{1_{[-A,A]}}K1-K1\|_\infty=\|1_{|x|>A}\|_\infty=1 $$ for all $A$.