Why the sample mean is not a good (sufficient) statistic? Consider a simple example of $X_{i}$ be i.i.d uniform distribution on the interval $[\theta,\theta+1]$. By strong low of large numbers, I may conclude that
$$\overline{X}\rightarrow_{P} \theta+\frac{1}{2} $$
However, it is not so clear what is the pdf or cdf of $\overline{X}$ as I would have to do an $n$-dimensional integral to get the cdf as $F_{\overline{X}}(t)=P(\sum X_{i}<nt,\theta\le X_{i}<\theta+1)$. Nevertheless it seems "intuitive" that $\overline{X}$ should be a good statistic. But practice one soon learns that the real sufficient statistic is $(X_{(1)},X_{(n)})$, and proving this via the factorization theorem is not difficult. 
I decided to ask the professor about this after the class. He told me that $\overline{X}$ is not a good statistic because it is not close enough to $\theta$, and as statisticans one has to consider real life applications. But this explanation is not persuasive, since I can use $\overline{X}-\frac{1}{2}$ for the same purpose. I want to ask if $\overline{X}$ is really a bad statistic for this example, and if yes for what reason. Also I want to know if $\overline{X}$ is a sufficient statistic. 
 A: Sufficiency pertains to data reduction, not estimation per se.  This is an important distinction to understand.  Yes, a "good" estimator is usually a function of a sufficient statistic, but that doesn't mean that all sufficient statistics are estimators.
As for your specific example, a simple way to understand why $\bar X$ is not a sufficient statistic for $\theta$ is to consider the following experiment:  suppose I tell you $\bar X = 10$.  Is this equivalent to all the information pertaining to $\theta$ that we can get from the sample?  Of course not:  for instance, $X_1 = 9.5, X_2 = 10.4, X_3 = 10.1$ could give us $10$, but so could $X_1 = X_2 = 9.75, X_3 = 10.5$.  If you only have knowledge of $\bar X$, you have lost information about $\theta$ that was available in the original sample:  namely, that in the first case, we must have $\theta \in [9.4,9.5]$, and in the second, $\theta \in [9.5,9.75]$.  Notice those intervals are nearly disjoint.  This is why $\bar X$ is insufficient for $\theta$.  You may be able to use it to estimate $\theta$ when $n$ is large, but as I have pointed out, sufficiency has to do with data reduction, not estimation.
