# What is the usual construction of Lebesgue measure?

So far, I've seen some different ways to construct Lebesgue measure $m$ on the Euclidean Space $\mathbb{R}^k$.

Are they all the same? And is $m$ a borel measure?

For instance, "Rudin-RCA" constucts Lebesgue measure by using Riesz Representation Theorem, that is $\sigma-$algebra $\mathfrak{M}$ defined in this way contains all the Borel sets of $\mathbb{R}^k$. That is, $\mathfrak{M}$ need NOT to be the set of all borel sets, hence $m$ is possibly somewhat stronger than Borel measure.

Are these Lebesgue measures actually the same? Or are they consistent? For example, i have learned two different definitions of Riemann-Stieltjes Integration and one of them is strictly stronger than another. Let's say $m_1$ is a Lebesgue measure constructed by using codable Borel sets and $m_2$ is the Lebesgue measure described above in the example. If a function $f$ is integrable with respect to $m_1$, then is $f$ integrable with respect to $m_2$?

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Yes these Lebesgue measures are the same as the Lebesgue measure is unique on $\mathbb R^k$. It is the unique translation invariant complete measure on the sigma algebra of Lebesgue measurable sets. The construction using Riesz is less common than the construction using Caratheodory's extension theorem. It seems to me that the Caratheodory construction is the most common. – Rudy the Reindeer Jan 24 '13 at 21:21
@Matt Is the domain of Lebesgue measure is exactly the completion of the set of all Borel sets of $\mathbb{R}^k$? And what is that uniqueness theorem called? – Katlus Jan 24 '13 at 21:24
My measure theory is shaky, at best. But I think the domain of the Lebesgue measure is the sigma algebra of Lebesgue measurable sets. As for the uniqueness: it follows from Caratheodory's extension theorem. – Rudy the Reindeer Jan 24 '13 at 21:32
Borel sets and Lebesgue measurable sets are not the same, if I recall correctly. – Jas Ter Jan 24 '13 at 21:37
There is a unique translation invariant measure on the Borel $\sigma$-algebra of $\mathbb{R}^k$ that asigns measure $1$ to the unit cube. The completion is called Lebesgue measure. Completions are always unique. Usualyy, one also calls the restriction to the Borel sets Lebesgue measure (I certainly do). But pretty much everyone agrees that a Lebesgue measurable set means a set in the completion of the $\sigma$-algebra. – Michael Greinecker Jan 25 '13 at 8:16