Uniform convergence and exchanging limit and integral I have to perform the following operation
$$\lim_{n\to\infty} \int_{-\infty}^{+\infty} \Phi(x+n) \phi(x) dx$$
where $\Phi$ and $\phi$ are the standard normal CDF and PDF respectively.
From what I understand, a sufficient condition to exchange limit and integral is that the integrand converges uniformly to its limit, which we know to be $\lim_{n\to\infty} \Phi(x+n) \phi(x) = \phi(x)$. Here, I don't think we have uniform convergence (but corrections are welcome as my understanding is still lacking), and one way to check it is to see that $\sup_x |\Phi(x+n)\phi(x)-\phi(x)| \not\rightarrow_{n\to\infty}0$.
However, solving the integral (using properties of standard normal integrals) yields
$$
\int_{-\infty}^{+\infty} \Phi(x+n) \phi(x) dx = \Phi(n)
$$
and so it turns out that
$$
\lim_{n\to\infty} \int_{-\infty}^{+\infty} \Phi(x+n) \phi(x) dx =  \int_{-\infty}^{+\infty} \lim_{n\to\infty} \Phi(x+n) \phi(x) dx = 1,
$$
so we can exchange limit and integral.
What am I missing? I am trying to understand the theory behind, because I will have to deal with more general functions than $\Phi(x+n)$, all of which have limits in this sort of additive way.
 A: If $f_n(x)\xrightarrow[n\to\infty]{}f(x)$ uniformly, then \begin{align*}
\left\lvert\int_a^bf_n(x)\,\mathrm dx-\int_a^bf(x)\,\mathrm dx\right\rvert&\le\int_a^b\lvert f_n(x)-f(x)\rvert\,\mathrm dx\\&\le\int_a^b\|f_n-f\|_\infty\,\mathrm dx\\&=(b-a)\|f_n-f\|_\infty\\&\xrightarrow[n\to\infty]{}0.\end{align*}
However, this is obviously assuming that $(a,b)$ is a finite interval, i.e., both $a$ and $b$ are finite.
Here you are integrating over $(-\infty,\infty)$. Nonetheless, you can use monotone convergence, as $(\Phi(\cdot+n))_{n\ge0}$ is non-decreasing (towards the constant function equals to $1$).
You can also use local uniform convergence as follows. On the one hand,
$$\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\le\int_{-\infty}^\infty\phi(x)\,\mathrm dx=1,$$
so $$\limsup_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\le1.$$
On the other hand, for each fixed $N>0$, by the uniform convergence $\Phi(x+n)\phi(x)\xrightarrow[n\to\infty]{}\phi(x)$ on $[-N,N]$:
$$\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-N}^N\Phi(x+n)\phi(x)\,\mathrm dx\xrightarrow[n\to\infty]{}\int_{-N}^N\phi(x)\,\mathrm dx,$$
which shows that
$$\liminf_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-N}^N\phi(x)\,\mathrm dx.$$
Letting now $N\to\infty$ leads to
$$\liminf_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx\ge\int_{-\infty}^\infty\phi(x)\,\mathrm dx=1.$$
Hence we have
$$\lim_{n\to\infty}\int_{-\infty}^\infty\Phi(x+n)\phi(x)\,\mathrm dx=1.$$
