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Show that if $m\left( E \right) <+\infty $ and $f:E\longrightarrow \left[ 0,+\infty \right] $ is measurable and the set $f^{ -1 }\left( \left( +\infty \right) \right) $ is of measure zero then for $\varepsilon >0,\exists M\in\mathbb{N} $ such that $m\left( \left\{ x:f\left( x \right)>M \right\} \right) <\varepsilon $. Proof: By contradiction

JAHP
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Put $E_M=\{x:f(x)\le M\}$. Since $\lim_{M\to\infty}E_M=\bigcup_{M>0}E_M=E\backslash\{x:f(x)=\infty\}$, $$ \lim_{M\to\infty}m(E_M)=m(E) $$ and therefore $m(\{x:f(x)>M\})=m(E)-m(E_M)$ goes to $0$ when $M\to\infty$.

Where is needed $m(E)<\infty$?

Veridian Dynamics
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    Or even easier, $f(x)=x$ on $\Bbb{R}$, then for all $M$, $m({x\colon f(x)>M})=\infty$ –  Jun 19 '17 at 04:45
  • What is limit of a sequence of sets? – edm Jun 19 '17 at 06:04
  • For non-decreasing sequences it is usually defined th elimit as $\lim_n A_n=\bigcup_{n\ge1}A_n$ $$.$$ See https://math.stackexchange.com/questions/234292/continuity-from-below-and-above for "continuity" properties of a measure. – Veridian Dynamics Jun 19 '17 at 12:50
  • Also, for "Where is needed $m(E)<\infty$ I mean "In the proof, where do I use the hypothesis $m(E)<\infty$. – Veridian Dynamics Jun 19 '17 at 12:52