Measures on the real line

Question

How do the conditions for a set to be measurable(inner -outer measures equality , approximately open measurable , complement sum measurable , Borel and measure 0 sigma algebra) change if the sets are subsets of $\mathbb{R}$ instead of $[0, 1]$?

Answer 1

I'm going to go ahead and assume that you've defined the Lebesgue outer-measure for subsets $E$ of $[0,1]$ according to the following: \begin{equation}\label{eq:1}\tag{1} m^\ast(E) := \inf_{\left\{ Q_k\right\}} \left\{ \sum_{k} |Q_k| : E \subseteq \bigcup_{k} Q_k\right\} \end{equation} where the $\{Q_k\}$ are countable families of intervals. It's also possible that you are restricting yourself to families of intervals contained in the set $[0,1]$. Then, as mentioned in the comments, a set $E \subseteq [0,1]$ is called Lebesgue measurable if for every $\epsilon > 0$ you can find an open set $O \supseteq E$ such that $m^\ast(O \setminus E) < \epsilon$.

Typically, this definition is used for all subsets of $\mathbb{R}$. Namely, we can define the outer-measure of any subset $E$ of $\mathbb{R}$ by simply using \eqref{eq:1} for $m^\ast(E)$. Of course, here we would have to consider intervals in $\mathbb{R}$ and not just in the interval $[0,1]$. We would also use the same definition of measurability: a set $E \subseteq [0,1]$ is called measurable if, for every $\epsilon > 0$, one can find an open set $O_\epsilon \supseteq E$ with $m^\ast(O_\epsilon \setminus E) < \epsilon$.

In what you call the "complement sum definition", we would instead define a set $E \subseteq \mathbb{R}$ to be measurable if $$ m^\ast(A) = m^\ast(A \cap E) + m^\ast(A \cap E^\complement) $$ for all subsets $A$ of $\mathbb{R}$. Here we would also be defining $m^\ast$ by taking the infimum (in \eqref{eq:1}) over families of intervals in $\mathbb{R}$ instead of $[0,1]$.