If $f \in L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$, the integral: $$\int_{\mathbb{R}^{n}} f(x) dx$$ is a well-defined object. If, however, $f \in L^{2}(\mathbb{R}^{n})$, we can define a different notion of the above integral by setting: $$\int_{\mathbb{R}^{n}}f(x)dx := \lim_{M \to \infty}\int_{|x| \le M}f(x)dx$$ provided the limit exists.
I have two questions concerning the above integral.
Q1: Is the second integral equivalent to the first one when restricted to $L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$, i.e. when $f \in L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$?
Q2: Suppose $K$ is an integral operator on $L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$, so that: $$(Kf)(x) := \int_{\mathbb{R}^{n}}K(x,y)f(y)dy$$ where I'm using the usual (first) integral. If $K_{1}$ and $K_{2}$ are both integral operators with integral kernels $K_{1}(x,y)$ and $K_{2}(x,y)$, then the operator $K := K_{1}K_{2}$ has integral kernel: $$K(x,y) = \int_{\mathbb{R}^{n}} K_{1}(x,z)K_{2}(z,y)dz$$ Does the same property hold when $L^{2}(\mathbb{R}^{n})\cap L^{1}(\mathbb{R}^{n})$ is replaced by $L^{2}(\mathbb{R}^{n})$ and the second notion of integral is used?