How can we study $\lim_{t\to\infty}\int_{\mathbb{R}}\psi_t(x)dx$, when $\lim$ and $\int$ do not commute? I am dealing with the problem of studying
$$\mathcal{I}= \lim_{t\to\infty}\int_{\mathbb{R}}\psi_t(x)dx$$
In my problem $\psi_t(x)$ is of the form $f_t(x)\mathbf{1}_{x>t}$. Moreover, $f_t(x)=\frac{g(x)}{g(t)}\varphi(t-x+k)$ where $g>
0$ is the probability density function of a continuous unbounded support random variable and $\varphi$ is the probability density function of a standard normal ($k$ being a constant).
It is clear that $\lim_{t\to\infty}\psi_t(x)$ is pointwise $0$. As a consequence, whenever I can bound $|\psi_t(x)|$ by an integrable function it's easy to conclude $\mathcal{I}=0$ by dominated convergence theorem. The ratio appearing in the integrand is always smaller than one (at least when $g$ is eventually decreasing) but, since the $f$ factor translates, it is difficult to find the desired bound.
Actually, depending on shape of $g$, this may not be possible.Indeed, numerical simulation suggest that, for many interesting choices of $g$, we have $\mathcal{I}\neq 0$. In these cases, it seems that it must not be possible to exchange integral and limit.
How to proceed ?
EDIT It is useful to notice that by de l'hospital $$-\lim_{t\to\infty}\frac{g'(t)}{g(t)}=\lim_{t\to\infty} \frac{g(t)}{1-G(t)}$$ which appears in the calculations below.
Here Computation of $ \lim_{t\to\infty}\int_t^\infty\frac{g(x)}{\sigma g(t)}f(\frac{t-x}{\sigma}+k)dx$ with $g$ and $\varphi$ density functions. I attempt a probably flawed calculation.
 A: Here's my go at this -- Severin anticipated my approach a bit.
One technical issue here is that $g$ is a valid pdf as long as $g>0$ almost everywhere (i.e., even if there is a subset  $X$ of Lebesgue measure $0$ where $g(x)=0\;\; \forall x \in X$).
Let's ignore that case and say that $g(x)>0 \;\;\forall x \in \mathbb{R}$.
Another edge case to ignore is where there exists a countable, increasing sequence $x_i$ such that $g(x_i)$ forms a divergent sequence. This will not matter in the integral but will complicate the limit proofs. See comments for some examples from others.

Expanding $\psi_i(x)$ to the full form (for clarity):
$$\psi_t(x) = \left[\frac{g(x)}{g(t)}\varphi(t-x+k)\right]\mathbf{1}_{x>t}$$
EDIT: As shown in the comments below, I cannot support the below solely based on $g \in L_1$, hence I will move it to an assumption:
Assume that $g$ eventually becomes a monotonically decreasing function bounded below by 0:
$$(*)\;\;\lim_{x\to -\infty} g(x) = \lim_{x \to +\infty} g(x)= 0$$
So we have:
$$\lim_{t \to \infty} \int_{\mathbb{R}} g(x)\mathbf{1}_{x>t} dx = 0$$
For any $t \in \mathbb{R}$ we also know
$$0 < \int_{\mathbb{R}} \psi_t(x)dx \; < \infty $$
Looking at the way you've defined $\psi_t(x)$, we see that the normal factor $\varphi$ translates along with the region where $\mathbf{1}_{x>t}$ is positive, so for all intents and purposes we can recast this by letting $\delta:= x-t$ (i.e., the distance into the non-zero part of the integral):
$$Q_t := \int_{0}^\infty \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta$$
We can see that for every $t$, $Q_t$ is basically an integral of a scaled and truncated version of $g$ multiplied by a gaussian with mean that is always located $k$ units to the right of $t$.
Now we can invoke $(*)$ to get what we need:
$$(*) \implies \forall t: \exists z_t>t: g(x) < g(t) \;\;\forall x>z_t$$
This means that we can decompose $Q_t$ into two parts (with $\Delta_t:=z_t-t$):
$$Q_t = \int_0^{\Delta_t} \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta + \int_{\Delta_t}^{\infty} \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta$$
Let's assume that that $\lim_{t\to \infty} \Delta_t =0$, which basically says that $g(x) \to h(x)$ where $h'(x) < 0\; \forall x$ is a monotonically decreasing function.
The first term is easy to evaluate now:
$$ \lim_{t\to \infty} \Delta_t = 0 \implies \int_0^{\Delta_t} \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta \to \int_0^{0} \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta = 0$$
But the second term becomes troublesome:
$$ 0 \leq \lim_{t\to\infty} \int_{\Delta_t}^{\infty} \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta \leq \lim_{t\to\infty}\int_{\Delta_t}^{\infty} \varphi(k-\delta)d\delta = \lim_{t\to \infty}-\Phi(k-t)+ \Phi(k-\Delta_t) = \frac{1}{2}$$
So you need a stronger condition than simply it eventually becoming monotonically decreasing.
Another bound we can examine is:
$$0 \leq \int_{\Delta_t}^{\infty} \frac{g(t+\delta)}{g(t)}\varphi(k-\delta)d\delta < \frac{1}{\sqrt{2 \pi}} \int_{\Delta_t}^{\infty} \frac{g(t+\delta)}{g(t)}d\delta$$
Pulling out the denominator we get:
$$\frac{1}{\sqrt{2 \pi}g(t)} \int_{\Delta_t}^{\infty} g(t+\delta)d\delta = \frac{1-F_g(t+\Delta_t)}{\sqrt{2 \pi}g(t)} $$
Taking the limit we have an indeterminate form:
$$ \lim_{t \to \infty} \frac{1-F_g(t+\Delta_t)}{\sqrt{2 \pi}g(t)} = \frac{0}{0}$$
Using L'Hospital's rule, we get:
$$ \lim_{t \to \infty} \frac{1-F_g(t+\Delta_t)}{\sqrt{2 \pi}g(t)} = \frac{-g(t+\Delta_t)}{\sqrt{2 \pi}g'(t)}$$
So here we get a sufficient condition, assuming $\Delta_t = 0$ without loss of generality (since if $\Delta_t$ is any increasing function of $t$ it just pushes $g$ towards its limit even faster).
$$ -g(t) = o(g'(t)) \implies \lim_{t \to \infty} \frac{-g(t+\Delta_t)}{\sqrt{2 \pi}g'(t)} = 0$$
So absolute rate of decrease of $g$ must become much larger than $g$ as $t\to\infty$
What kind of functions satisfy this? It can't be the exponential:
$$g(x) \propto e^{-x} \implies \frac{-g(x)}{g'(x)} = 1$$
EDIT: I stand corrected! The Weibull with shape > 1 does indeed satisfy the above condition and so is a valid $g$ for this problem (as found numerically by OP). If we let scale param be 1 (without loss of generality), we get these results:
$$ \frac{-g(x)}{g'(x)}=\frac{-k x^{k - 1} e^{-x^k}}{(-ke^{-x^k}) x^{k - 2}(k(x^k - 1) + 1)} = \frac{ x}{kx^k + 1-k}$$
$$k \in (0,1) \implies \lim_{x\to \infty} \frac{ x}{kx^k + 1-k}= \infty$$
$$k = 1 \implies \lim_{x\to \infty} \frac{ x}{kx^k + 1-k}=1$$
$$k > 1 \implies \lim_{x\to \infty} \frac{ x}{kx^k + 1-k}=0$$
