# Show a Set $A$ is Not the Countable Union of $0$ Content Sets where $A^c$ is Meagre

Background: This is a delicate one, so bear with me for a little

We know the set of discontinuity points of a Riemann integrable has $$0$$ measure since it is a countable union of $$0$$ content sets, where I define a $$0$$ content set as follows:

A set $$X\subset \mathbb{R}$$ has $$0$$ content if $$\forall \epsilon>0; \enspace k \in \mathbb{N}$$, finite

• $$X \subset \bigcup\limits_{i=1}^{k} (a_i,b_i)$$ and also $$\sum_{i=1}^k (b_i - a_i) < \epsilon$$

For my definition of 0 measure, the only difference is this union and sum can be infinite.

Now, a natural question would be if every $$X\subset \mathbb{R}$$ where X has measure $$0$$ can be written as the countable union of content $$0$$ sets. I know the answer is no, since I have a problem that is asking me to show that a specific set, A, defined bellow, is exactly a counterexample of this statement.

Problem

Let A be a set of $$0$$ measure such that its complement, $$A^c$$ is a countable union of closed sets with empty interior. ($$A^c= \bigcup\limits_{n} C_n$$)

Show that $$A$$ can't be written as the countable union of $$0$$ content sets ($$A \ne \bigcup\limits_{n} B_n$$)

My Reasoning

Since we know $$A^c= \bigcup\limits_{n} C_n$$ where $$c_n$$ has empty interior, $$A^c$$ has to have empty interior because of Baire Theorem. But if we take the complement of that, we conclude $$A$$ is dense in $$\mathbb{R}$$. This also means the closure of A, (cl(A)) is $$\mathbb{R}$$

Now we assume, by contradiction, that $$A$$ can be written as the countable union of $$0$$ content sets ($$A = \bigcup\limits_{n} B_n)$$. If we take the calourure on both sides: $$cl(A)=\mathbb{R}=cl(\bigcup\limits_{n} B_n) \supset \bigcup\limits_{n} cl(B_n))$$

But notice that, since $$B_n$$ has content $$0$$, it has empty interior and so does $$cl(B_n)$$. Using Baire theorem once again, $$\bigcup\limits_{n} cl(B_n)$$ has empty interior

This gives us nothing

because there is no contradiction in saying there is an empty interior set in $$\mathbb{R}$$, so I don't know what else to do.

A zero content set neccessarily has empty interior. This follows from subadditivity of measure/content. Moreover, the closure of a zero content set also has zero content. We can prove this as follows: suppose $$X$$ has zero content, write $$X \subset \bigcup\limits_{i=1}^{k} (a_i,b_i)$$ and $$\sum_{i=1}^k (b_i - a_i) < \epsilon$$. Taking closures on both sides, we get that $$cl(X)\subset \bigcup\limits_{i=1}^{k} [a_i,b_i]$$. By widening each interval by $$\frac{\epsilon}{k}$$, we obtain that $$cl(X)\subset \bigcup\limits_{i=1}^{k} (a_i-\frac{\epsilon}{2k},b_i+\frac{\epsilon}{2k})$$, with $$\sum_{i=1}^k [(b_i+\frac{\epsilon}{2k})-(a_i-\frac{\epsilon}{2k})] \leq 2\epsilon$$. As this holds for each $$\epsilon >0$$, it follows that $$cl(X)$$ also has zero content. Now we can prove your problem: Suppose that $$A=\bigcup_n B_n$$ with each $$B_n$$ a zero content set, and $$A^c = \bigcup_n C_n$$ with each $$C_n$$ closed and having empty interior. Then, $$A\subset \bigcup_n cl(B_n)$$, where each set $$cl(B_n)$$ is closed and has zero content, hence empty interior by the above reasoning. It follows that $$\mathbb{R} = cl(A) \cup A^c = \bigcup_n (cl(B_n) \cup C_n)$$ which contradicts the Baire Category Theorem.