This was a quiz question which I did not get full credit for.


Let $\epsilon$>0 be given.

Let $A= \{x_1,...,x_k\}$

Let $I_j = (x_j - \frac{\epsilon}{4k}, x_j + \frac{\epsilon}{4k})$ , $j=1,....,k$

and $\sum^{j=1}_{k} m(I_j) = \frac{\epsilon}{2}$ Hence measure zero

Then $m(I_j)=\frac{\epsilon}{2k}$, $A \subset \cup^{j=1}_{k}I_j$

Let $B= \{y_1,...,y_l\}$

Let $I_p = (x_p - \frac{\epsilon}{4l}, x_p + \frac{\epsilon}{4l})$ , $p=1,....,l$

Then $m(I_p)=\frac{\epsilon}{2l}$, $B \subset \cup^{p=1}_{l}I_p$

and $\sum^{p=1}_{l} m(I_p) = \frac{\epsilon}{2}$ Hence measure zero

Thus $A\cup B \leq \sum m(I_j) + \sum m(I_p) = \epsilon$

My professor commented why do I assume A,B is finite? (I took this assumption because I saw an example problem with measure zero but I lack explanation why it was finite). He then states not proven, not shown. He mentioned proving using discontinuities which I could not figure out.

  • 5
    $\begingroup$ measure zero set could have infinite points. You can simply use the subadditivity of measure to prove it. $\endgroup$ – hermes Mar 5 '16 at 6:55

To expand hermes's comment: what if $A = \mathbb{N} \subseteq \mathbb{R}$? Your proof fails at the second line.

It is one of the defining properties of a measure $\mu$ that if $A, B$ are disjoint then $$\mu(A \cup B) = \mu(A) + \mu(B)$$

Therefore, consider the two sets $A, B$ as three disjoint sets: $$A \setminus (A \cap B), B \setminus (A \cap B), A \cap B$$

They are all subsets of measure-zero sets so they all have measure zero (since $\mu(A \setminus (A \cap B)) + \mu(A \cap B) = 0$, for instance); so their total $\mu$ is $0$.


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