This is my previous attempt: Show that the union of two sets of measure zero, has a measure zero
Need help with my flow of logic, and accuracy.
Proof
Assume $f, g$ be Riemann Integrable on $[a,b]$.
$\forall \epsilon>0$
Let $u$ be the set of discontinuities of $f$ and $v$ be the set of discontinuities of $g$.
Since u and v have measure zero, by defintion $\exists I_i = \{I_1,I_2,….\}$ s.t.
$u,v \subset \cup_{i=1}^{\infty} I_i$ and $\sum_{i=1}^{\infty}m(I_i) < \epsilon$
Since $u$ and $v$ are mutually disjoint,
$m(u \cup v) = m(u) + m(v)$
Then for, $u \setminus (u \cap v)), u \cap v \subset m(u \cup v)$
We have $m(u \setminus (u \cap v))) + m(u \cap v) = 0$
By definition, $u\cup v \subset \cup_{i=1}^{\infty} I_i$ and $\sum_{i=1}^{\infty}m(I_i) < \epsilon$
Hence the union of the two sets has measure zero.
Best Answer
I think you can just use the definitions here:
Let $A$ and $B$ be sets of Lebesgue measure zero, so that there are sequences $\left \{ I _k\right \}_{k\in \mathbb N}\ $and $\left \{ J _k\right \}_{k\in \mathbb N}\ $ s.t $A\subseteq \bigcup _k I_k$ and $B\subseteq \bigcup _k J_k$ and s.t $\ \forall \epsilon >0,$
$\sum _k\vert I_k\vert <\epsilon$ and $\sum _k\vert J_k\vert <\epsilon$. Now, since $\left \{ I_k,J_{k'} \right \}_{k,k'}$ is countable, arrange it in some way to obtain a countable sequence $\left \{ I'_n \right \}_n$, which has the desired properties:
$A\cup B\subseteq \bigcup _nI'_n$
and
$\sum _n\vert I'_n\vert <2\epsilon$