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Need to connect Borel sets and sigma algebra. Also, need to show that a lebesgue measure contains all of the Borel Sets. I know that borel sets are the smallest open sigma algebra sets but I need to further my understanding and was having trouble with this.

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  • $\begingroup$ Could you be more specific? On $R^n$ with finite n>0, the set of Lebesgue sets can be defined as the sigma-algebra generated by the union of A and B, where B is the Borel sets, and x belongs to A iff x is a subset of a measure-zero Borel set. It follows that y is a Lebesgue set iff $p\subset y\subset q$ where p is G-delta, and q is F-sigma ,and p,q have equal measure,. Also it follows that a subset of a Lebesgue-null set is Lebesgue-null. $\endgroup$ – DanielWainfleet Dec 8 '15 at 5:20
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Let $X \neq \varnothing$; let $\mathscr{X} \subset 2^{X}$. Then by proof the set $$ \sigma (\mathscr{X}) := \bigcap \{ \mathscr{Y} \subset 2^{X} \mid \mathscr{Y} \supset \mathscr{X} \ \text{is a}\ \sigma\text{-algebra} \} $$ is a $\sigma$-algebra; call $\sigma(\mathscr{X})$ the $\sigma$-algebra generated by $\mathscr{X}$.

Let $(X, \mathscr{X})$ be a topological space; let $E \subset X$. Then $E$ is called a Borel set iff $E \in \sigma (\mathscr{X})$.

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