Showing that ∫ X log ⁡ ( f ) d μ ≤ μ ( X ) log ⁡ 1 μ ( X ) given that ∫ X f d μ = 1 for positive f without using Jensen's inequality.We weren't presented with Jensen's inequality, but with Fubini's Theorem and another theorem which is:in a finite measure space f ∈ L 0 + ( Σ ) satisfies ∫ X f d μ = ∫ ( 0 , ∞ ) μ ( { f &gt; t } d m ( t ) )At first I thought I should use { | f | &lt; 1 μ ( X ) }, { | f | ≥ 1 μ ( X ) } and show that if ∫ log ⁡ f d μ &gt; μ ( X ) log ⁡ ( 1 μ ( X ) ), then ∫ log ⁡ f d μ ≤ μ ( { | f | &lt; 1 μ ( X ) } ) log ⁡ 1 μ ( X ) + ∫ { | f | ≥ 1 μ ( X ) } log ⁡ f d μ and therefore log ⁡ 1 μ ( X ) ( μ ( X ) − μ ( { | f | &lt; 1 μ ( X ) } ) ) = log ⁡ 1 μ ( X ) ( μ ( { | f | ≥ 1 μ ( X ) } ) ) &lt; ∫ μ ( { | f | ≥ } ) log ⁡ f d μwhich I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality?

Question

Showing that       ∫          X        log  ⁡  (  f  )  d  μ  ≤  μ  (  X  )  log  ⁡            1              μ        (        X        )             given that       ∫          X        f  d  μ  =  1 for positive   f without using Jensen&#039;s inequality.We weren&#039;t presented with Jensen&#039;s inequality, but with Fubini&#039;s Theorem and another theorem which is:in a finite measure space   f  ∈      L          0              +        (  Σ  ) satisfies       ∫          X            f    d    μ    =      ∫          (      0      ,      ∞      )        μ  (  {  f  &amp;gt;  t  }  d  m  (  t  )  )At first I thought I should use   {      |    f      |    &amp;lt;            1              μ        (        X        )              },  {      |    f      |    ≥            1              μ        (        X        )              } and show that if   ∫  log  ⁡  f  d  μ  &amp;gt;  μ  (  X  )  log  ⁡  (            1              μ        (        X        )              ), then  ∫  log  ⁡      f    d  μ  ≤  μ  (  {      |    f      |    &amp;lt;            1              μ        (        X        )              }  )  log  ⁡            1              μ        (        X        )              +      ∫          {              |            f              |            ≥                        1                      μ            (            X            )                              }        log  ⁡      f    d  μ and therefore  log  ⁡            1              μ        (        X        )              (  μ  (  X  )  −  μ  (  {      |    f      |    &amp;lt;            1              μ        (        X        )              }  )  )  =  log  ⁡                    1                  μ          (          X          )                      (  μ  (  {      |    f      |    ≥            1              μ        (        X        )              }  )  )  &amp;lt;      ∫          μ      (      {              |            f              |            ≥      }      )        log  ⁡      f    d  μwhich I can&#039;t seem to contradict. Any ideas how this can be shown, without using Jensen&#039;s Inequality?

hopeloothab9m · Accepted Answer

Use that  log  ⁡  (  x  )  ≤  log  ⁡  (  s  )  −  1  +      x    s  for any constant   s  &amp;gt;  0. Then take   s  =  μ  (  X      )          −      1       and use this bound to get an upperbound for the integrand and hence also for the integral. (This is, all said and done, just Jensen’s inequality.)

Showing that ∫ <mrow class="MJX-TeXAtom-ORD"> X </mrow>

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