Finding likelihood given uniform data I'm using this past paper to pre-study for a module in Bayesian Statistics. It has a question,

A precision weighing device yields unbiased measurements within half a gramme,
which can be modelled as $\text{Un}(x|\theta-1/2,\theta+1/2)$, where $\theta$ is the unknown weight. A priori, it is believed $\theta\sim\text{Un}(10,20)$. Using $\textbf{x} = \{11, 11.5, 11.7, 11.1, 11.4, 10.9\}$, a set of six independent measurements, find the posterior distribution of $\theta$.

I'd like to reason that
\begin{align}
f(\textbf{x}|\theta)
& = \prod_{i=1}^6f(x_i|\theta) \\
& = \begin{cases}
\prod_{i=1}^6\frac{1}{(\theta+1/2)-(\theta-1/2)} = 1 & \text{if } \theta \in \bigcap_{i=1}^6[x_i-1/2,x_i+1/2] = [11.2,11.4], \\
0 & \text{otherwise;}
\end{cases}
\end{align}
and then if the prior and the normalising constant happen to cancel we'd have something like $\theta|\textbf{x}\sim\text{Un}(11.2,11.4)$; but I think this $f(\textbf{x}|\theta)$ only integrates to $0.2$, so isn't a pdf. Could someone point me in the right direction?
Edit: On reflection, I now think the relationship $f(\textbf{x}) = \prod_{i=1}^nf(x_i)$ only holds for ordered data, e.g. $$p((H,T))=p((T,H))=0.5\times0.5=0.25\neq0.5=p((H,T))+p((T,H))=p(\{H,T\}).$$ But eliminating ordering by dividing by $n!$ seems like overkill in the weighing example.
 A: The likelihood $f(\mathbf x| \theta)$ for fixed $\mathbf x = (11,11.5,11.7,11.1,11.4,10.9)$ is a function in $\theta$ but not a density in $\theta$, so it doesn't have to integrate to one.
On the other hand, you are looking for the posterior
$$
f(\theta|\mathbf x) = \frac{f(\mathbf x|\theta)f(\theta)}{f(\mathbf x)} 
$$
which is a density in $\theta$, so it does have to integrate to one.
Your original computation of $f(\mathbf x|\theta)$ as the indicator function
$$
I_{[11.2, 11.4]}(\theta) = \left\{\begin{matrix}
                                   1 & \mbox{for}\;\theta\in [11.2, 11.4]\\
                                   0 & \mbox{otherwise}\\
                                   \end{matrix}\right.
$$
is correct, and since the support $[11.2, 11.4]$ is a subset of the support of the prior, the nominator of the posterior is the same as $f(\mathbf x|\theta)$. Then, since we have said above that the posterior does have to integrate to one, we deduce that the evidence $f(\mathbf x)$ has to be 0.2, resulting in the posterior being
$$
f(\theta|\mathbf x) = 5\cdot I_{[11.2, 11.4]}(\theta).
$$
