probability distributions - Interchange of limit and integral for a positive random variable with finite moments.

Let $f$ be the pdf of a non-negative random variable $X$ with finite moments of all orders, i.e. $E[X^n]<+\infty$ for all $n\in \mathbb N$. May I interchange the limit with the integral and infer that $$\lim_{w\to \infty}\int_{0}^{+\infty}xf(x+w)\mathop{dx}=\int_{0}^{+\infty}x\cdot\lim_{w\to \infty}f(x+w)\mathop{dx}=0$$

Answer

To elaborate on @Did's comments, by a change of variables we see that
$$\mathbb E[(X-w)^+] = \int_0^\infty (x-w)^+f(x)\mathsf dx = \int_0^\infty xf(x+w)\mathsf dx. $$
Since $X\geqslant0$, $0\leqslant(X-w)^+\leqslant X$ for all $w\geqslant 0$, and as $X\in\mathcal L^1$, by dominated convergence we have that
$$\lim_{w\to\infty}\mathbb E[(X-w)^+] = \mathbb E\left[\lim_{w\to\infty}(X-w)^+\right].$$

Since $\mathbb E[X]<\infty$, for any $\omega$, we may choose $w$ such that $w>X(\omega)$, and hence $$\lim_{w\to\infty}(X-w)^+=0.$$

It follows that

$$\lim_{w\to\infty}\int_0^\infty xf(x+w)\mathsf dx = \mathbb E\left[\lim_{w\to\infty}(X-w)^+\right] = 0.$$