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For $n \in \mathbb{N}$, and for $X_{(n)} = \max\left(X_1, \ldots, X_n\right)$ where $X_1, \ldots, X_n \sim Unif(0,1)$ iid. I have that:

$$ P\left(\int_{X_{(n)}}^{1}\frac{1}{\theta^n}e^{-(\theta-1)^{-2}}d\theta>C\right) $$

for a $C >0$. I would ultimately like to show that

$$ \lim_{n \to \infty}P\left(\int_{X_{(n)}}^{1}\frac{1}{\theta^n}e^{-(\theta-1)^{-2}}d\theta>C\right) = 0 $$

However, I am having trouble trying to separate out the stochastic element, $X_{(n)}$ off to one side. Does anyone see anything I can do here? Thanks.

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    $X_{(n)}\to 1$ with probability $1$. So, the integral tends to $0$; becomes greater than any $C<1$. Is this not enough? – 2017-02-14
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    $X_{(n)}\approx 1-1/n$, so that integral is something like $e^{-n^2}$, converging to $0$ very fast. – 2017-02-18

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We know that $X_{(n)}\rightarrow 1$ in probability. We also know that convergence in probability is conserved under continuous functions, so $f(X_{(n)})\rightarrow f(1)$ for any continuous function $f$ (in fact we only need continuity at $1$). As $f(x)=\int_x^1\frac{1}{\theta^n}e^{-(\theta-1)^{-2}}d\theta$ is continuous at 1 we have that $\mathbb{P}[f(X_{(n)})>C]\rightarrow \mathbb{P}[f(1)>C]=\mathbb{P}[0>C]=1_{C<0}$.