Let f be measurable in R and ∫ R f d x = 1 Show that for any r ∈ ( 0 , 1 ), there is a Lebesgue measurable set E such that ∫ E f d x = r.I have tried the interval ( − R , R ). If I can show that R ↦ ∫ − R R f d x is continuous and I can use the intermediate value theorem to show that there is a R such that ∫ − R R f d x = r. If I consider | ∫ f ( χ ( − R 1 , R 1 ) − χ ( − R 2 , R 2 ) ) d x | , I do not know to show the continuous since I think that ∫ − R R f d x may not be finite for some R or ∫ | f | d x may not be finite.

Question

Let   f be measurable in       R   and       ∫          R        f  d  x  =  1 Show that for any   r  ∈  (  0  ,  1  ), there is a Lebesgue measurable set   E such that       ∫    E    f  d  x  =  r.I have tried the interval   (  −  R  ,  R  ). If I can show that   R  ↦      ∫          −      R        R    f  d  x is continuous and I can use the intermediate value theorem to show that there is a   R such that       ∫          −      R        R    f  d  x  =  r. If I consider       |    ∫  f  (      χ          (      −              R        1            ,              R        1            )        −      χ          (      −              R        2            ,              R        2            )        )  d  x      |  , I do not know to show the continuous since I think that       ∫          −      R        R    f  d  x may not be finite for some   R or   ∫      |    f      |    d  x may not be finite.

Jovany Mathews · Accepted Answer

So   f  ∈      L          1        (            R        ), so the absolute continuity of integral implies that given that   ϵ  &amp;gt;  0, there is some   δ  &amp;gt;  0 such that             ∫              F                    |        f    (    x    )          |        d    x    &amp;lt;    ϵ   for any measurable set   F with       |    F      |    &amp;lt;  δ.Now Lebesgue Dominated Convergence Theorem implies that             ∫              [        −        R        ,        R        ]              f    (    x    )    d    x    →    1   since       χ          [      −      R      ,      R      ]            |    f      |    ≤      |    f      |  .And we have             ∫              [        −        R        ,        R        ]              f    (    x    )    d    x    →    0   as   R  ↓  0, so such an   α is guaranteed by Intermediate Value Theorem.

alexmjn · Accepted Answer

As an addition   ∫  f  d  x  =  1 imply that   f  ∈      L    1   because in Lebesgue integration,       ∫    f   exists if and only if       ∫          |        f          |       exists.

Let f be measurable in R and int_R fdx=1 Show that for any r in(0,1), there is a Lebesgue measurable set E such that int_E fdx=r.

Open question

Answer & Explanation

New Questions in Calculus 2