passing an integral to the inside of a min 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? 
 A: The proposition is false without additional assumptions on $f_n$.
Counterexample
Let $h$ be the hat function on $[0,1]$, i.e. the function
$$ h(x) = \max\left( 1 - |2x-1|,0 \right).$$ 
Notice that $h$ is zero outside of $[0,1]$ and $\int h(x) \,d x=1.$ 
Let $f_n$ be the negative translate of $h$, i.e.
$$ f_n(x)=-h(x-n)$$ 
Since those translates have disjoint support (i.e. they do not overlap) $\min_n f_n = \sum_n f_n.$ Now if you take the $\min$ over $N$ of those translates
$$ \min_n \int f_n(x)\, dx = \min_n (-1) = -1.$$
but $$ \int \min_n f_n (x) \, dx= \int \sum_n f_n(x) \, dx = -N.$$ 
A: It is true that if you have two functions $f,g$ with $f \le g$, then $\int f \le \int g$. Would this not imply equality if there truly is a $\min$ of the functions? However, if you were to replace the minimum with $\inf$, I am unsure whether the equality still holds.
Edit: I can expound. If we let $f_0 = \min f_n$, then it is true that $$\int \min_{k\in \mathbb{N}} f_k=\int f_0 \le \int f_n$$ for all $n\in \mathbb{N}$. So $$\int f_0 = \min_{n\in \mathbb{N}} \int f_n$$ Thus implying that $$\int \min_{n\in \mathbb{N}} f_n = \min_{n\in \mathbb{N}} \int f_n$$
