Let ( Ω , A , ν ) be a measure space. Show that if the measure over the space Ω is 0 &lt; ν ( Ω ) &lt; ∞, then Q = 1 ν ( Ω ) νis a probability measure.I don't see how this is indeed a probability measure. For it to be a probability measure indeed it has to1. Return values in [0,1].2. Satisfy countable additivity.3. Implied measure of ∅ should be 0.The only thing I currently know is that ν is a measure and that it is finite (and greater than 0) over Ω. This one seems quite tough.

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Let   (  Ω  ,      A    ,  ν  ) be a measure space. Show that if the measure over the space   Ω is   0  &amp;lt;  ν  (  Ω  )  &amp;lt;  ∞, then  Q  =      1          ν      (      Ω      )        νis a probability measure.I don&#039;t see how this is indeed a probability measure. For it to be a probability measure indeed it has to1. Return values in [0,1].2. Satisfy countable additivity.3. Implied measure of   ∅ should be 0.The only thing I currently know is that   ν is a measure and that it is finite (and greater than 0) over   Ω. This one seems quite tough.

cuprins60 · Accepted Answer

It&#039;s not difficult. View that&amp;nbsp;Q&amp;nbsp;(&amp;nbsp;Ω&amp;nbsp;)&amp;nbsp;=&amp;nbsp;ν&amp;nbsp;(&amp;nbsp;Ω&amp;nbsp;)&amp;nbsp;ν&amp;nbsp;(&amp;nbsp;Ω&amp;nbsp;)&amp;nbsp;=&amp;nbsp;1, and your first criterion is met by monotonicity. It is evident that countable additivity results from the countable additivity of&amp;nbsp;ν&amp;nbsp;directly, andQ&amp;nbsp;(&amp;nbsp;∅&amp;nbsp;)&amp;nbsp;=&amp;nbsp;ν&amp;nbsp;(&amp;nbsp;∅&amp;nbsp;)&amp;nbsp;ν&amp;nbsp;(&amp;nbsp;Ω&amp;nbsp;)&amp;nbsp;=&amp;nbsp;0&amp;nbsp;ν&amp;nbsp;(&amp;nbsp;Ω&amp;nbsp;)&amp;nbsp;=&amp;nbsp;0

Let ( <mi mathvariant="normal">Ω , <mrow class="MJX-TeXAtom-ORD">

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