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: 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.
