# any $\sigma $-finite measure is equivalent to a probability measure

The following theorem states that for any $\sigma $-finite (http://planetmath.org/SigmaFinite) measure^{} $\mu $, there is an equivalent^{} probability measure $\mathbb{P}$ — that is, the sets $A$ satisfying $\mu (A)=0$ are the same as those satisfying $\mathbb{P}(A)=0$.
This result allows statements about probability measures to be generalized to arbitrary $\sigma $-finite measures.

###### Theorem.

Any nonzero $\sigma $-finite measure $\mu $ on a measurable space^{} $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{)}$ is equivalent to a probability measure $\mathrm{P}$ on $\mathrm{(}X\mathrm{,}\mathrm{A}\mathrm{)}$. In particular, there is a positive measurable function^{} $f\mathrm{:}X\mathrm{\to}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty}\mathrm{)}$ satisfying $\mathrm{\int}f\mathit{}\mathit{d}\mu \mathrm{=}\mathrm{1}$, and $\mathrm{P}\mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}{\mathrm{\int}}_{A}f\mathit{}\mathit{d}\mu $ for all $A\mathrm{\in}\mathrm{A}$.

###### Proof.

Let ${A}_{1},{A}_{2},\mathrm{\dots}$ be a sequence in $\mathcal{A}$ such that $$ and ${\bigcup}_{k}{A}_{k}=X$. Then it is easily verified that

$$g\equiv \sum _{k=1}^{\mathrm{\infty}}{2}^{-k}\frac{{1}_{{A}_{k}}}{1+\mu ({A}_{k})}$$ |

satisfies $1\ge g>0$ and $$. So, setting $f=g/\int g\mathit{d}\mu $, we have $\int f\mathit{d}\mu =1$ and therefore $\mathbb{P}(A)\equiv {\int}_{A}f\mathit{d}\mu $ is a probability measure equivalent to $\mu $. ∎

Title | any $\sigma $-finite measure is equivalent to a probability measure |
---|---|

Canonical name | AnysigmafiniteMeasureIsEquivalentToAProbabilityMeasure |

Date of creation | 2013-03-22 18:33:44 |

Last modified on | 2013-03-22 18:33:44 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 6 |

Author | gel (22282) |

Entry type | Theorem |

Classification | msc 28A12 |

Classification | msc 28A10 |

Related topic | SigmaFinite |

Related topic | Measure |