I am reading the first few pages of Dembo&Zeitouni. I have some questions about the alternative definition of the large deviation principle in page 6. The equation (1.2.7) states that

For every $\alpha < \infty$ and every measurable set $\Gamma$ with $\bar{\Gamma} \subset \Psi_I(\alpha)^c$, $$\limsup_{\epsilon \to 0} \epsilon \log\mu_{\epsilon} (\Gamma) \leq -\alpha.$$

is equivalent to the definition to the LDP upper bound $$\limsup_{\epsilon \to 0} \epsilon \log\mu_{\epsilon} (\Gamma) \leq -\inf_{x\in \bar{\Gamma}} I(x) $$.

It is not clear to me that why they are equivalent.One direction is obvious. However, why the former implies the latter is not clear to me. I tried to take $\alpha = \inf_{x\in \bar{\Gamma}} I(x)$, however, I can not show $\bar{\Gamma}$ is a subset of $\Psi_I(\alpha)^c$.

Can anyone shed some light on this? Thank you in advance.

  • $\begingroup$ Suppose for a moment that $\Gamma $ is closed. Then $I>\alpha $ on $\Gamma $. Do you see the implication in this case? Now think about the semicontinuity of $I $. $\endgroup$ – Ian Nov 6 '15 at 17:16
  • $\begingroup$ @Ian Do you still take $\alpha = \inf_{x\in \bar{\Gamma}} I(x)$ in this case? When $\Gamma$ is closed, it is still possible that there exists some $x\in \Gamma$ such that $I(x) = \alpha$ $\endgroup$ – ting Nov 6 '15 at 19:57
  • $\begingroup$ If $\Gamma $ is not closed then you will still have $I \geq \alpha $ there because of the semicontinuity. $\endgroup$ – Ian Nov 6 '15 at 20:19
  • $\begingroup$ @Ian thanks, Ian. But I am still a little confused. For any $\Gamma$, we take $\alpha = \inf_{x \in \bar{\Gamma}} I(x)$. If we can show that $\bar{\Gamma} \subset \Psi_I(\alpha)^c$ (which is equivalent to showing for any $x \in \bar{\Gamma}$, $I(x)$ is strictly bigger than $\alpha$), then we are done. However, by the definition of $\alpha$, we only have $I(x) \geq \alpha$. My question is how we rule out the case $I(x) = \alpha$? $\endgroup$ – ting Nov 7 '15 at 0:00
  • $\begingroup$ You can't necessarily take $\alpha$ to be equal to the infimum, because you don't know that $\overline{\Gamma}$ will then be contained in $\Psi(\alpha)^c$, because maybe the infimum is actually attained. But whether it is attained or not, you can still apply the hypothesis with $\alpha = \inf_{x \in \overline{\Gamma}} I(x) - \varepsilon$ for arbitrarily small positive $\varepsilon$. $\endgroup$ – Ian Nov 7 '15 at 0:03

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