I was wondering when I could do something like the following:
$$\int \min\limits_{n\in\mathbb{N}}f_n = \min_{n\in \mathbb{N}} \int f_n$$ when I don't necessarily have a decreasing sequence of functions.
The problem I have is the following:
Imagine we have some really nice random variables $(X_i)_{i=1}^n \stackrel{iid}{\sim} X$ in the sense that they are $L^1$ and their cdfs are bijective and continuous and perhaps some other properties that maybe I need. Then I want to show that
$$\min_{\theta \in \mathbb{R}} \bigg( \theta + \frac{1}{n(1-\epsilon)} \sum_{i=1}^n \max (-X_i - \theta, 0)\bigg)$$ is an unbiased estimator for
$$\min_{\theta \in \mathbb{R}} \bigg( \theta + \frac{1}{(1-\epsilon)} \mathbb{E}[\max (-X - \theta, 0)] \bigg)$$
I would like to just pass the Expectation from the outside to the inside, but I'm having some trouble showing how to do this. It has been a while since I've taken real analysis.
So far, it seems that for each $\omega \in \Omega$,
$$f_{\theta}(\omega) := \theta + \frac{1}{n(1-\epsilon)} \sum_{i=1}^n \max (-X_i(\omega) - \theta, 0)$$ is continuous and convex in $\theta$ with negative derivative as $\theta \rightarrow -\infty$ and positive derivative as $\theta \rightarrow \infty$. So then for each $\omega \in \Omega$,
$$\inf_{\theta \in \mathbb{R}} f_{\theta}(\omega) = \min_{\theta \in \mathbb{R}} f_{\theta}(\omega)$$
and
$$\min_{\theta \in \mathbb{R}} f_{\theta}(\omega) = \min_{\theta \in \mathbb{Q}} f_{\theta} (\omega)$$
Hence
$$ \min_{\theta \in \mathbb{R}} f_{\theta}$$ is measurable.
All I can think to use would be the Dominated Convergence Theorem, but I'm not sure if I can come up with any sequence that converges to my estimator pointwise which is also bounded in $L^1$.
Does anyone have any advice?