TITLE: What mistake is in my "proof" that the set of all subsets of R with lebesgue measure 1 has a maximal element?
QUESTION [6 upvotes]: I apologize in advance if the mistake is trivial. But I'm simply not seeing it. I must have misunderstood something along the way. The idea follows:
Let $K = \{E \subset \mathbb{R} | m(E) = 1\}$, where m is the Lebesgue measure.
Now, let's define the following order in $K$. We say that $A < B$ iff $B \subseteq A$. This is obviously a partial order.
Now, let $(A_i)$ be any chain. (i.e. $A_i < A_{i+1}$)
Let $A = \bigcap A_i$. Clearly $A > A_i$ for all $i$.
Also $A_i \searrow A$, and $m(A_i) < \infty$ for all i. Then applying a Corollary we have that
$m(A) = \lim_{n\rightarrow \infty} m(A_i) = 1$, and therefore $A \in K$
Therefore, every chain in $K$ has an upper bound in $K$.
Using Zorn's Lemma, we conclude that $K$ has a maximal element. But this conclusion seems absolutely absurd to me. In fact, take any candidate $X$ to being a maximal element. Clearly $X$ is non empty. Let $x \in X$. Now $X - \{x\} \in K$ and $X < X - \{x\}$.
I believe that my mistake is that, to be able to use the corollary, it is necessary that $A$ be measurable.
What went wrong here?
Thank you very much!

REPLY [4 votes]: A "chain" is just any totally ordered subset of $K$.  It doesn't have to be a sequence indexed by the natural numbers; it might be uncountable.  So you can't assert that $m(A) = \lim_{i\rightarrow \infty} m(A_i) = 1$, since you don't necessarily have a sequence at all.