For $Y \sim Beta(n, 1)$, showing that $\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx$ converges in probability to $-\infty$? Suppose that $Y \sim Beta(n, 1)$. I am interested in showing that 
$$
\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx
$$
converges in probability to $-\infty$. (I was told in a book this is true)
My set-up is to show for $K>0$:
$$
\lim_{n \to \infty}P\left(\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx
<-K\right) = 1
$$
Now,
\begin{align}
P\left(\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx
<-K\right) &= P\left(\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx
<e^{-nK}\right)\\
&\leq P\left(\int_{Y}^1 x^{-1}e^{-(x-1)^{-2}}dx
<e^{-nK}\right) \\
& \leq P\left(\int_{Y}^1 1\cdot dx
<e^{-nK}\right) \\
&= P\left(Y > 1-e^{-nK}\right) \\
&= 1-\left(1-e^{-nK}\right)^n
\end{align}
Above, my 2nd to 3rd inequality was obtained as $x^{-n}e^{-(x-1)^{-2}}$ is bounded below by $x^{-1}e^{-(x-1)^{-2}}$, hence the terms inside the probability for $x^{-n}e^{-(x-1)^{-2}}$ is a subset of the terms inside the probability for $x^{-1}e^{-(x-1)^{-2}}$.
The same for my 3rd to 4th inequality.
Applying the limit, the entire probability term goes to $0$. Hence:
$$
\lim_{n \to \infty}P\left(\frac{1}{n}log\int_{Y}^1 x^{-n}e^{-(x-1)^{-2}}dx
<-K\right) = 0
$$
which is different from the result above. Does anyone know where I might have messed up? Thanks.
 A: Let $\{Y_n\}_{n\geq1}$ a sequence of random variables defined on the same probability space such that $Y_n\sim Beta(n,1)$ and $Y_n<Y_{n+1}$
we can see that that $g(u)=\frac{1}{n}log\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$ is a concave function, so by Jensen's inequality $E(g(Y_n))\leq g(E(Y_n))$ and $\{g(Y_n)\}_{n\geq1} $ is a decreasing sequence lower than zero, thus convergent. We have that
$$g(E(Y_n))=g\left(\frac{n}{n+1}\right)=\frac{1}{n}log\int_{\frac{n}{n+1}}^1 x^{-n}e^{-(x-1)^{-2}}dx$$
which converges to $-\infty$ as $n\to \infty. $ and so does $g(Y_n)\to-\infty$, and since the sequence converges, it converges almost surely and probability.
The problem in your demonstration is, as stated by NCh, that $\begin{align}P\left(\int_{Y}^{1}x^{-1}e^{-(x-1)^{-2}}dx<e^{-nK}\right) & \leq P\left(\int_{Y}^{1}1\cdot dx<e^{-nK}\right)\end{align}$ is not true.
Proof of $\frac{1}{n}log\int_{\frac{n}{n+1}}^1 x^{-n}e^{-(x-1)^{-2}}dx\to-\infty$:
Using L'hôpital, if $\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx \to 0$
$$\begin{align*}
\lim_{n\to\infty}\frac{log\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx}{n} &=\lim_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^{n}e^{-\left(n+1\right)^{2}}/\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx}{1}\\
&=\lim_{n\to\infty}\frac{\left(1+\frac{1}{n}\right)^{n}e^{-\left(n+1\right)^{2}}}{\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx}\\
 &=e\lim_{n\to\infty}\frac{-2\left(n+1\right)e^{-\left(n+1\right)^{2}}}{\left(1+\frac{1}{n}\right)^{n}e^{-(n+1)^{2}}}\\
 &=\lim_{n\to\infty}-2\left(n+1\right)\\
 &=-\infty\end{align*}
$$
Proof of $\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx \to 0$:
$\begin{align*}\lim_{n\to\infty}\int_{\frac{n}{n+1}}^{1}x^{-n}e^{-(x-1)^{-2}}dx &\leq\lim_{n\to\infty}\int_{\frac{n}{n+1}}^{1}\left(\frac{n}{n+1}\right)^{-n}e^{-(\frac{n}{n+1}-1)^{-2}}dx\\
 &=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{-n}e^{-(\frac{n}{n+1}-1)^{-2}}\left(1-\frac{n}{n+1}\right)\\
 &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}e^{-(n+1)^{2}}\left(\frac{1}{n+1}\right)\\
 &=0\end{align*}$
Proof of $g(u)=\frac{1}{n}log\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$ being concave:
let f,g be concave twice differentiable functions, if f is increasing then
$$f\left(h\left(u\right)\right)'' =f''\left(h\left(u\right)\right)h'\left(u\right)^{2}+f'\left(h\left(u\right)\right)h''\left(u\right)\leq0$$
if we apply that first to
$f(u)=u^{\frac{1}{n}}$, $h(u)=\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$
and then to
 $f(u)=log(u)$ and $h(u)=\left(\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx\right)^{\frac{1}{n}}$
Proof of $h(u)=\int_{u}^1 x^{-n}e^{-(x-1)^{-2}}dx$ being concave:
$$
\begin{align*}
h'(u) & =u^{-n}e^{-(u-1)^{-2}}\\
h''(u) & =-nu^{-n}e^{-(u-1)^{-2}}-2u^{-n}e^{-(u-1)^{-2}}(1-u)^{-3}
\end{align*}
$$
