Question: > 
> If $\mu(E)=0$, show that $\mu(E\cup A)=\mu(A\setminus E)=\mu(A)$.
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I just started learning about measure this week, so I don't know any theory about measure except the definition of outer measure and that a countable set has measure zero.

Remark: $\mu(\cdot)$ denoted the outer measure

Attempt.

Suppose $\mu(E)=0$. Then $\mu(A\cup E)\leq\mu(A)+\mu(E)=\mu(A)$.

If I can also show that $\mu(A \cup E)\geq \mu (A)$, then equality would follow.

I know that if $A$ and $E$ were disjoint compact sets, then $\mu(A)+\mu(E)\leq \mu(A \cup E)$, so maybe I can say something like a compact set in $\mathbb{R}$ is closed and bounded by Heine-Borel, and somehow use the fact that $\mu([a,b])=\mu((a,b))$ to help.

For $\mu(A\setminus E)=\mu(A)-\mu(E)=\mu(A)$

A hint in the right direction would be appreciated. Thanks.

Answer: 
If $A,B$ are disjoint we have $\mu(A \cup B) = \mu A + \mu B$.

If $A \subset B$ we have $\mu A \le \mu B$.

We have $A = (A \setminus E) \cup (A \cap E)$, and the latter two sets are
disjoint so
$\mu A = \mu (A \setminus E) + \mu (A \cap E)$.
Note that $\mu(A \cap E) \le \mu E$ and $A \setminus E \subset A$.

Now note that $A \cup E = (A \setminus E) \cup E$, and the latter two sets are
disjoint.
Another way is to use the definition 

$$\mu(A\cup E)=\mu((A\cup E)\cap E)+\mu((A\cup E)\setminus E)=\mu(E)+\mu(A\setminus E)=\mu(A\setminus E)$$

and 

$$\mu(A)=\mu(A\cap E)+\mu(A\setminus E)=\mu(A\setminus E).$$