Let ( X , A , μ ) be a measure space and we consider a measurable positive function f : X → [ 0 , + ∞ ]. I already proved that if the Lebesgue integral of f on X is finite, that is ∫ X f d μ &lt; + ∞ ,then μ ( { f = + ∞ } ) = 0, that is f is finite almost everywhere.Now, let f : X → [ − ∞ , + ∞ ] an μ- integrable function, that is ∫ X f + d μ &lt; + ∞ ∫ X f − d μ &lt; + ∞ ,then ∫ X | f | d μ = ∫ X ( f + + f − ) d μ &lt; ∞ ,and therefore | f | is finite ae, so is a measurable positive function.From this can I conclude that f itself is finite almost everywhere?Notation. f + := max { f ( x ) , 0 } and f − := max { − f ( x ) , 0 }

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Let   (  X  ,      A    ,  μ  ) be a measure space and we consider a measurable positive function   f  :  X  →  [  0  ,  +  ∞  ]. I already proved that if the Lebesgue integral of   f on   X is finite, that is      ∫    X    f    d  μ  &amp;lt;  +  ∞  ,then   μ  (  {  f  =  +  ∞  }  )  =  0, that is   f is finite almost everywhere.Now, let   f  :  X  →  [  −  ∞  ,  +  ∞  ] an   μ- integrable function, that is      ∫    X        f    +      d  μ  &amp;lt;  +  ∞        ∫    X        f    −      d  μ  &amp;lt;  +  ∞  ,then      ∫    X        |    f      |      d  μ  =      ∫    X    (      f    +    +      f    −    )    d  μ  &amp;lt;  ∞  ,and therefore       |    f      |   is finite ae, so is a measurable positive function.From this can I conclude that   f itself is finite almost everywhere?Notation.      f    +    :=  max  {  f  (  x  )  ,  0  } and       f    −    :=  max  {  −  f  (  x  )  ,  0  }

Melina Glover · Accepted Answer

Yes, if&amp;nbsp;|&amp;nbsp;f&amp;nbsp;|&amp;nbsp;if a.e. is finite, then likewise&amp;nbsp;f, because any location&amp;nbsp;f&amp;nbsp;is infinite would force&amp;nbsp;|&amp;nbsp;f&amp;nbsp;|&amp;nbsp;to be infinite as well.

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