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The measurement of Lebesgue owes its name to the French mathematician Henri Leon Lebesgue. It is of an major importance in theory of integration.

Formal definition

That is to say $(\backslash \; R,\; B\; (\backslash \; R))$ the measurable Space $\backslash \; R$ provided with its Tribe borélienne. There exists single a measurement noted $\backslash \; lambda$ on this measurable Espace which has the two following properties:

1. $\backslash \; forall\; has\; \backslash \; in\; \backslash \; R,\; \backslash \; forall\; has\; \backslash \; in\; B\; (\backslash \; R),\; \backslash \; lambda\; (has\; +\; A)=\; \backslash \; lambda\; (A)$ (invariance by translation)
2. $\backslash \; lambda\; ()\; =1\; \backslash ,$.

This measurement is called measurement of Lebesgue on $\backslash \; R$. Moreover, one can show that it coincides with the concept length on the intervals, i.e. the measurement of Lebesgue of an interval is equal to the length of this interval: for example, $\backslash \; lambda\; (1;\; 9)\; =9-1=8\; \backslash ,$. Same manner, $\backslash \; lambda\; (]\; -4;\; 8])\; =\; 8\; -\; (-\; 4)\; =\; 12\; \backslash ,$. It is not EC way which Lebesgue introduced historically this measurement.

Note:: if $has\; \backslash \; in\; \backslash \; R$ and $A\; \backslash \; subset\; \backslash \; R$, one noted $a\; +\; has\; \backslash ,$ the unit: $\backslash \; \{has\; +\; X,\; X\; \backslash \; in\; has\; \backslash \}$

Properties of the measurement of Lebesgue

• the measurement of Lebesgue is finished on the limited parts of $\backslash \; R$, in particular on the compact ones, which means that:

In particular, . And since $\backslash \; R$ is equal to the union of all the boréliens $\backslash ,$ when N traverses $\backslash \; N$, one says that the measurement of Lebesgue is $\backslash \; sigma$-finie.

• the measurement of Lebesgue is outside regular , which means that:

$\backslash \; forall\; has\; \backslash \; in\; B\; (\backslash \; R),\; \backslash \; lambda\; (A)=\; \backslash \; inf\; \backslash \; \{\backslash \; lambda\; (O),\; O\; \backslash \; text\; \{open\; of\}\; \backslash \; R,\; \backslash ,\; has\; \backslash \; subset\; O\; \backslash \}$.

• the measurement of Lebesgue is internally regular , which means that:

$\backslash \; forall\; has\; \backslash \; in\; B\; (\backslash \; R),\; \backslash \; lambda\; (A)=\; \backslash \; sup\; \backslash \; \{\backslash \; lambda\; (K),\; K\; \backslash \; text\; \{compact\; of\}\; \backslash \; R,\; \backslash ,\; K\; \backslash \; subset\; has\; \backslash \}$.

Tribe of Lebesgue

One has just seen that the measurement of Lebesgue is a measurement on the Tribu borélienne of $\backslash \; R$. However, this tribe is not largest on which one can define this measurement.

Together negligible for the measurement of Lebesgue

That is to say $N\; \backslash ,$ part of $\backslash \; R$. It is said that $N\; \backslash ,$ is a together negligible for the measurement of Lebesgue if there exists a borélien $A\; \backslash \; in\; B\; (\backslash \; R)$ such as:

1. $N\; \backslash \; subset\; A$
2. $\backslash \; lambda\; (A)=0\; \backslash \; quad\; \backslash ,$

The negligible parts of $\backslash \; R$ are thus the units included in a borélien of null measurement of Lebesgue. One notes $N\_\; \{\backslash \; lambda\}\; \backslash ,$ the whole of the negligible parts of $\backslash \; R$.

Definition of the tribe of Lebesgue

By definition, the tribe of Lebesgue on $\backslash \; R$, noted $L\; (\backslash \; R)$, is the tribe generated by the union of $B\; (\backslash \; R)$ and of $N\_\; \{\backslash \; lambda\}\; \backslash ,$. One shows in fact that it is equal to:

$L\; (\backslash \; R)\; =\; \backslash \; sigma\; (\backslash \; \{has\; \backslash \; cup\; NR,\; has\; \backslash \; in\; B\; (\backslash \; R),\; NR\; \backslash \; in\; N\_\; \{\backslash \; lambda\}\; \backslash \})$

The tribe of Lebesgue is thus the tribe generated by the units which are written like the union of a borélien and of a negligible unit. Since the empty set is of null measurement of Lebesgue and that the union of a borélien with the empty set is equal to this same borélien, it results from it that the tribe of Lebesgue contains the tribe borélienne.

As opposed to what one could think, the tribe of Lebesgue is not equal to $P\; (\backslash \; R)$ (the whole of the parts of $\backslash \; R$) and this result is obtained thanks to the use of the Axiome of the choice. In other words, there exists part of $\backslash \; R$ which is not in the tribe of Lebesgue; to see Together not-measurable.

Extension of the measurement of Lebesgue

Now that one defined the tribe of Lebesgue, one can see that one can extend the measurement of Lebesgue on this tribe so that new measurement obtained coincides with the measurement of Lebesgue on the boréliens.

One poses, for all borélien $A\; \backslash ,$ and for any negligible unit $N\; \backslash ,$:

$\backslash \; Lambda\; (\backslash \; cup\; NR\; has)\; =\; \backslash \; lambda\; (a)$.

One can show that $\backslash \; Lambda$ is well defined and that it is a measurement on the tribe of Lebesgue.

Space measured $(\backslash \; R,\; L\; (\backslash \; R),\; \backslash \; Lambda)$ is then what is called a complete space measured , which means that it contains all its negligible units. In other words, if $N$ is part of $\backslash \; R$ such as it exists $B\; \backslash \; in\; L\; (\backslash \; R)$ with $\backslash \; Lambda\; (B)=0\; \backslash \; quad$ and $N\; \backslash \; subset\; B$ then $N\; \backslash \; in\; L\; (\backslash \; R)$. It is also said that the tribe of Lebesgue is the supplemented tribe of the tribe borélienne for the measurement of Lebesgue.

Cardinal of the tribe of Lebesgue

Intuitively, it is felt well that the tribe of Lebesgue on $\backslash \; R$ is much larger than the Tribu borélienne. One proves that rigorously by showing that:

1. $\backslash \; mbox\; \{card\}\; \backslash ,\; (L\; (\backslash \; R))\; =\; \backslash \; mbox\; \{card\}\; \backslash ,\; (P\; (\backslash \; R))$
2. $\backslash \; mbox\; \{card\}\; \backslash ,\; (B\; (\backslash \; R))\; =\; \backslash \; mbox\; \{card\}\; \backslash ,\; (\backslash \; R)$

Known as differently, that means that the tribe of Lebesgue is in bijection with the whole of the parts of $\backslash \; R$ (although it is not equal for him when it is supposed that the axiom of the choice is true) whereas the tribe borélienne is simply in bijection with $\backslash \; R$. However it is well-known in Set theory that $\backslash \; R$ is not in bijection with $P\; (\backslash \; R)$ (and more generally, that is true for any unit). Consequently, one will be able to never find a bijection between the tribe of Lebesgue and the tribe borélienne, which wants to say well that the tribe of Lebesgue contains more elements than the tribe of Borel. It results that one with following strict inclusion:

$B\; (\backslash \; R)\; \backslash \; subsetneq\; L\; (\backslash \; R)$

Measure of Lebesgue on $\backslash \; R^\; \{N\}$

One considers now the measurable Espace $(\backslash \; R^\; \{N\},\; B\; (\backslash \; R^\; \{N\}))$, i.e. space $\backslash \; R^\; \{N\}$ provided with its Tribe borélienne. One will see that one can define the measurement of Lebesgue on this space. More generally, one can define measurements of Lebesgue on the Euclidean vector spaces .

Theorem-definition

There exists single a measurement on space $(\backslash \; R^\; \{N\},\; B\; (\backslash \; R^\; \{N\}))$, that one will note $\backslash \; lambda\_\; \{N\}\; \backslash ,$ such as:

1. $\backslash \; forall\; has\; \backslash \; in\; \backslash \; R^\; \{N\},\; \backslash \; forall\; has\; \backslash \; in\; B\; (\backslash \; R^\; \{N\}),\; \backslash \; lambda\_\; \{N\}\; (a+A)=\; \backslash \; lambda\_\; \{N\}\; (A)\; \backslash ,$ (invariance by translation)
2. $\backslash \; lambda\_\; \{N\}\; (^\; \{N\})\; =1\; \backslash ,$

This measurement is called measurement of Lebesgue on $(\backslash \; R^\; \{N\},\; B\; (\backslash \; R^\; \{N\}))$.

Explanations

Let us take the case of $\backslash \; R^\; \{2\}$. The measurement of Lebesgue $\backslash \; lambda\_\; \{2\}\; \backslash ,$ on this space coincides on the rectangles of the form $\backslash \; times\; \backslash ,$ with the concept of surface of those. Indeed, one proves that one has $\backslash \; lambda\_\; \{2\}\; (\backslash \; times)\; =\; (Ba)\; (cd.)\; \backslash ,$. More generally, the measurement of Lebesgue $\backslash \; lambda\_\; \{2\}\; \backslash ,$ of a subset borélien of $\backslash \; R^\; \{2\}$ corresponds to our intuitive definition of the surface: for example, the measurement of Lebesgue of a disc of ray $a\; \backslash ,$ is equal to: $\backslash \; pi.\; a^\; \{2\}\; \backslash ,$. Same manner, if one considers space $\backslash \; R^\; \{3\}$, the measurement of Lebesgue $\backslash \; lambda\_\; \{3\}\; \backslash ,$ on this space corresponds to our intuitive défintion of the volume, and it is thus without surprise that the measurement of Lebesgue of a ball of ray $a\; \backslash ,$ is worth $\backslash \; frac\; \{4\}\; \{3\}.\; \backslash \; pi.\; a^\; \{3\}$.

See too

• Measurement;
• Tribe and Tribe borélienne;
• Together of null measurement;
• Integral of Lebesgue;
• Paradox of Banach-Tarski;
• secondary Measurements.

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