# limit of the integral $\lim_{N \to \infty} \int_b^{\infty} f_N(x) x dx = 0$

Consider a sequence of functions $$f_N(x)$$ for which I know $$\limsup_{N \to \infty} \frac{1}{N} \log f_N(x) = -g(x)$$, where $$g(x)$$ is positive, continuous on $$(a, \infty)$$, and increases to infinity as $$x \to \infty$$. Moreover, $$$$g(x) \geq - \log(x)+V(x)+C \hspace{1cm} \text{for } x \geq a, \text{ and } C \text{ a positive constant}$$$$ where $$V(x)$$ is a continuous function satisfying $$$$\liminf_{x \to \infty} \frac{V(x)}{\beta \log(x) } >1 \hspace{1cm} \text{for } \beta >1$$$$

I want to show that $$\lim_{N \to \infty} \int_b^{\infty} f_N(x) x dx = 0$$ for $$b >a$$.

$$\bullet Origin$$ of the problem comes from that $$f_N$$'s are probability measures, ($$P(X_n >x)$$), satisfying a large deviation principle with good rate function $$g(x)$$ with the above properties, and I want to show that $$\mathbb{E} [ X_n \mathbb{I}\{ X_n > b \} ]$$ converges to 0 as $$N \to \infty$$.

My first question is that, what is the interpretation of $$\liminf_{x \to \infty} \frac{V(x)}{\beta \log(x) } >1$$? can we deduce that $$V(x) > \beta \log(x)$$?

Assuming $$V(x) > \beta \log(x)$$ for all $$x$$, here is what I have tried so far:

First note that we have $$g(x) \geq \beta' \log(x) + C$$ for positive constants $$\beta', C$$. From $$\limsup_{N \to \infty} \frac{1}{N} \log f_N(x) = -g(x)$$, given $$0<\epsilon , for large enough $$N$$,we have $$f_N(x) \leq e^{-N(g(x) - \epsilon)}$$. So, we have: $$\int_b^{\infty} f_N(x) x dx \leq \int_b^{\infty} e^{-N(g(x) - \epsilon)} x dx \leq e^{-N(C-\epsilon)} \int_b^{\infty} e^{-N \beta' \log(x)} x dx = e^{-N(C-\epsilon)} \frac {b^{2 - N \beta'}}{N \beta' -2} \hspace{5mm} \text{for sufficiently large} N$$ taking the limit, we have the result.

I'm not sure if it is correct or not. Any help is appreciated!

Without uniformity on $$x$$, it is not necessary true. Consider $$f_N(x) = \begin{cases} \exp(-N\cdot g(x)), x < N\\ \text{anything}, x \geq N \end{cases}$$
Then not just $$\liminf$$, but even $$\lim\limits_{N\to \infty} \frac{1}{N} f_N(x) = -g(x)$$, but $$\int_b^\infty f_N(x)\, dx$$ can be made arbitrary for each $$N$$ independently.
Assuming uniformity in the first limit ($$\forall \varepsilon > 0 \exists N_0 \forall x: \frac{1}{N}\log f_N(x) < -g(x) + \varepsilon$$) - that's what needed for $$f_N(x) \leq e^{-N(g(x) - \epsilon)}$$ you are almost correct.
$$\liminf\limits_{x \to \infty} h(x) > 1$$ means exactly that for some $$\varepsilon > 0$$ and $$x_0$$, $$h(x) > 1 + \varepsilon$$ if $$x > x_0$$. So you have $$g(x) \geq \beta' \log(x) + C$$ if $$x > x_0$$. Then your reasoning shows that $$\int_{x_0}^\infty f_N(x)\, dx \to 0$$.
We only need to deal with $$\int_b^{x_0}f_N(x)\, dx$$. But $$g(x)$$ is bounded from $$0$$ on $$[b, x_0]$$ (as continuous positive function on closed segment), so let's say $$\inf_{x \in [b, x_0]} g(x) = \delta > 0$$, and as for large enough $$N$$, $$F_N(x) \leq \exp(-N(g(x) - \delta / 2)) \leq \exp(-N \delta / 2)$$, for large enough $$N$$, $$\int_{b}^{x_0} f_N(X)\, dx \leq \exp(-N \delta / 2) \cdot (x_0 - b) \to 0$$.