Let ( X , A , μ ) be a measure-space and f : X → [ 0 , ∞ ] a unsigned measurable map. Show that μ ( { f ( x ) &gt; 0 } ) = 0 ⟺ μ ( { f ( x ) &gt; t } ) = 0 For every t &gt; 0.Suppose that μ ( { f ( x ) &gt; 0 } ) = 0. Since f = 0 a.e we have that ∫ X f = 0. Thus μ ( { f ( x ) &gt; t } ) ≤ 1 t ∫ X f = 0 and since the measure is positive we have that μ ( { f ( x ) &gt; t } ) = 0.Conversely suppose that μ ( { f ( x ) &gt; t } ) = 0. Then what can be done for this case? It seems that I could write ∫ X f = ∫ X ∖ { f ( x ) &gt; t } f + ∫ { f ( x ) &gt; t } f and the latter integral would be zero, but I don’t really get anywhere from here?

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Let   (  X  ,      A    ,  μ  ) be a measure-space and   f  :  X  →      [    0  ,  ∞  ] a unsigned measurable map. Show that   μ  (  {  f  (  x  )  &amp;gt;  0  }  )  =  0    ⟺    μ  (  {  f  (  x  )  &amp;gt;  t  }  )  =  0 For every   t  &amp;gt;  0.Suppose that   μ  (  {  f  (  x  )  &amp;gt;  0  }  )  =  0. Since   f  =  0 a.e we have that       ∫    X    f  =  0. Thus   μ  (  {  f  (  x  )  &amp;gt;  t  }  )  ≤      1    t        ∫    X    f  =  0 and since the measure is positive we have that   μ  (  {  f  (  x  )  &amp;gt;  t  }  )  =  0.Conversely suppose that   μ  (  {  f  (  x  )  &amp;gt;  t  }  )  =  0. Then what can be done for this case? It seems that I could write       ∫    X    f  =      ∫          X      ∖      {      f      (      x      )      &amp;gt;      t      }        f  +      ∫          {      f      (      x      )      &amp;gt;      t      }        f and the latter integral would be zero, but I don’t really get anywhere from here?

grindweg1v · Accepted Answer

Note that   {  f  (  x  )  &amp;gt;  0  }  =      ⋃          n      ≥      1        {  f  (  x  )  &amp;gt;  1      /    n  }, it holds  μ  (  {  f  (  x  )  &amp;gt;  0  }  )  =      lim          n      →      ∞        μ  (  {  f  (  x  )  &amp;gt;  1      /    n  }  )  =  0since the sequence of set       A    n    =  {  f  (  x  )  &amp;gt;  1      /    n  } is increasing.

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