MH644 ASSIGNMENT NO. 4 SPRING 2022 SOLUTION

MTH644

ASSIGNMENT NO. 4

SOLUION:-

Problem:

Let,

$A \subseteq R$ such that m(A) = 0. If A is bounded then show that the set $B = \left\{ {{x^2}|x \in A} \right\}$ is lebesgue measurable and m(B) = 0.

Let,

$A \subseteq R$ such that m(A) = 0.

If A is bounded we have to show that $B = \left\{ {{x^2}|x \in A} \right\}$ is lebesgue measurement and m(B) = 0.

For,

Lebesgue measure

Any set $E \subseteq m$ is defined to be lebesgue outer measure.

${m^ * } = m$

For a class A,

${m^ * }\left( A \right) = {m^ * }\left( {A \cap F} \right) + {m^ * }\left( {A \cap {F^ * }} \right)$

For non – empty measurable set.

$f:E \to \left[ { - \infty ,\infty } \right]$

$= \left\{ {x|x \in E:f\left( x \right) > c} \right\} \in m\left( R \right)\forall c \in R$

$f:E \to \left[ { - \infty ,\infty } \right]\,\,as\,\,B = f\left( x \right) = {x^2}\forall x \in R$

$\left\{ {x \in R:f\left( x \right) > c} \right\} = \left\{ {x \in R:{x^2} \ge c} \right\}$

$\left( { - \infty ,c} \right) \cup \left( {c,\infty } \right) \in m\left( R \right)$

$\left\{ {x \in R:{x^2} > 1} \right\} = \left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$

$\,f$ is lebesgue measure.

MH644 ASSIGNMENT NO. 4 SPRING 2022 SOLUTION

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