'Easy' question on continuity of integral Here's a problem that I recently stumbled upon. It seems pretty easy, and quite intuitive yet every time I try to solve it, I run into some difficulties. Here it goes :
Let $\phi \in BC(\mathbb{R}^2,\mathbb{R})$ (bounded and continuous) be a function such that
$$\forall_{x \in \mathbb{R}} \ \phi(x, \cdot) \in L_1(\mathbb{R},\mathbb{R})$$
Then the function defined as 
$$\psi(x) = \int_{\mathbb{R}} \ |\phi(x,z)| \ dz $$
is continuous. 
I got as far as to trying to estimate the expression 
$$\int_{\mathbb{R}\backslash [-T,T]} \ |\phi(x,z) - \phi(x_0,z)| \ dz$$
for large $T$ (we want the expression to be small). However, the main problem is that $T$ may depend on $x$, in other words it need not be uniform for $x$ in some neighbourhood of $x_0$. I feel like it would somehow violate the continuity, but maybe I am totally wrong.. I would be grateful for providing a simple counterexample if the statement is false. 

Here is a counterexample:
$$|\phi(x,z)| = x^2e^{-x}|z|e^{-x|z|}\chi_{[0,\infty)}(x).$$
The function is bounded and continuous, but
$$\psi(0) = 0,$$
and for all $x > 0$,
$$\psi(x) = x^2e^{-x}\int_{-\infty}^{\infty}|z|e^{-x|z|}\,dz= 2x^2e^{-x}\int_{0}^{\infty}ze^{-xz}\,dz=2e^{-x}>0.$$
