Processing math: 74%
ThmDex – An index of mathematical definitions, results, and conjectures.
Lebesgue outer measure under scaling
Formulation 0
Let Rd be a D816: Euclidean real Cartesian product.
Let μ be the D780: Lebesgue outer measure on Rd.
Let λR{0}.
Then ERd:μ(λE)=|λ|dμ(E)
Proofs
Proof 0
Let Rd be a D816: Euclidean real Cartesian product.
Let μ be the D780: Lebesgue outer measure on Rd.
Let λR{0}.
Let ERd. Let {In}nN be a countable covering for E of open real d-intervals. Result R2987: Set covering of Euclidean real scaled set shows than then the collection of scaled intervals {λIn}nN is in turn a countable covering for λE of open real d-intervals. Results
(i) R1049: Lebesgue outer measure is outer measure
(ii) R1052: Lebesgue outer measure coincides with the volume function on real n-intervals
(iii) R1159: Euclidean volume function under scaling

now imply μ(λE)μ(λnNIn)nNμ(λIn)=nNVol(λIn)=|λ|dnNVol(In) Since {In}nN was an arbitrary covering of E, we may apply R1109: Antitonicity of infimum to extend the above inequality to the infimum element on the right hand side over all such coverings of E to obtain μ(λE)|λ|dinf We may now proceed to apply this inequality in turn to the scaled set \frac{1}{\lambda} E to obtain an inequality in the opposing direction \begin{equation} \mu^*(E) = \mu^* \Big( \lambda \Big( \frac{1}{\lambda} E \Big) \Big) = \mu^* \Big( \frac{1}{\lambda} (\lambda E) \Big) \leq \frac{1}{|\lambda|^d} \mu^*(\lambda E) \end{equation} Multiplying both sides by the nonzero quantity |\lambda|^d yields now |\lambda|^d \mu^*(E) \leq \mu^*(\lambda E). An inequality in both directions was established, whence R1043: Equality from two inequalities for real numbers implies the equality \mu^*(\lambda E) = |\lambda|^d \mu^*(E). This finishes the proof. \square