Minimal sufficiency with indicator functions The following theorem can be used to demonstrate that a statistic is minimal sufficient:

Let $f(X|\theta)$ be the pmf or pdf of a sample X. Suppose $\exists$ a
  function $T(X)$ such that, for every two samples X and Y:
$\frac{f(X|\theta)}{f(Y|\theta)} \amalg \theta$ iff $T(X) = T(Y)$. 
Then $T(X)$ is a minimal suffienct statistic for $\theta$.

(Reference: Casella & Berger Theorem 6.2.13.) 
I'm confused about how this works when the densities contain indicator functions. Here is an example that I constructed in order to confuse myself:
Let $X_{i}$ and $Y_{i} \sim Unif(0,\theta)$. Let $I$ be the indicator function. Then:
$$
\frac{f(X|\theta)}{f(Y|\theta)} = \frac{ I( max(X) < \theta ) \cdot I( min(X) > 0 ) }{ I( max(Y) < \theta ) \cdot I( min(Y) > 0 )}$$
We'll forget about the ratio of the second indicator functions because it's already independent of $\theta$. So the question becomes, "When is the ratio of the first indicator terms independent of $\theta$?" The ratio will be 1 if $I( max(X) < \theta ) = I( max(Y) < \theta ) = 1$. If $I( max(X) < \theta ) = I( max(Y) < \theta ) = 0$, I'm hazy on what to do because the denominator is 0, but at least the ratio shouldn't depend on $\theta$. Okay -- so $max(X)$ is a candidate MSS, and it fulfills the "if" direction.
Let's look at the "only if" direction. If $max(X) \ne max(Y)$, then the ratio is either $1/0$ or $0/1$. To my mind, these are independent of $\theta$ regardless of any function of $X$ or $Y$. But this conclusion doesn't make any sense; therefore the premise that I understand indicator functions and minimal sufficiency is wrong.
I'd love some guidance to set my understanding straight.
 A: Someone described to me a clever proof sketch that I will flesh out for any interested parties.
First, let's fix the problem about having 0 in the denominator by constructing an equivalent condition:
$$
\frac{f(X|\theta)}{f(Y|\theta)} = \frac{ I( max(X) < \theta ) }{ I( max(Y) < \theta )} $$
is equivalent to:
$$ \exists \; g(X,Y) \amalg \theta \; \text{such that:}$$
$$ \; {I( max(X) < \theta ) = g(X,Y) \cdot I( max(Y) < \theta )} \; \forall \; \theta \; \; \; \; \; \; (*)$$
Take $max(X)$ as the candidate MSS. For the "if" direction, simply note that when $max(X) = max(Y)$, $(*)$ holds regardless of $\theta$, if we let $g(X,Y) = 1$.
For the "only if" direction, we need to show that $(*)$ implies $max(X) = max(Y)$. 
We will prove the contrapositive: $max(X) \ne max(Y)$ implies $(*)$ does not hold, i.e., $g(X,Y)$ depends on $\theta$. WLOG, let $max(X) < max(Y)$.
Now there are three cases:
(1) $\theta < max(X) < max(Y)$. Then both indicators are 0, so $g(X,Y)$ can be anything and $(*)$ will hold.
(2) $max(X) < \theta < max(Y)$. Then the indicators don't match, so $g(X,Y)$ must evaluate to 0 to fulfill $(*)$. 
(3) $max(X) < max(Y) < \theta$. Then the indicators are both 1, so $g(X,Y)$ must evaluate to 1 to fulfill $(*)$. 
Therefore, $max(X) \ne max(Y)$ implies that $(*)$ does not hold, since $g(X,Y)$ would be a piecewise function dependent on $\theta$. And we're done.
