Question

Suppose $f$ is a measurable function and $f(x) > 0$ for all $x$. Let $g(x) = \frac{1}{f(x)}$. Prove that g is a measurable function.

score 828 · Answer 1 · 2017-10-02 22:46:47

828

Views

Suppose $f$ is a measurable function and $f(x) > 0$ for all $x$. Let $g(x) = \frac{1}{f(x)}$. Prove that g is a measurable function.

Published on 02 Oct 2017 - 22:46

score 734 · Answer 2 · 2017-10-03 18:25:09

Suppose $(X, \mathcal{A})$ is a measurable space, $f$ is a real valued function, and $\{x : f(x) > r\} \in \mathcal{A} $ for each rational number $r$.

score 2.8k · Answer 3 · 2017-10-03 19:51:08

2.8k

Views

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue measurable, prove that there exists a Borel measurable function $g$ s.t. $f = g$ a.e.

Published on 03 Oct 2017 - 19:51

score 248 · Answer 4 · 2017-10-06 17:56:03

248

Views

Embedding of some function spaces

Published on 06 Oct 2017 - 17:56

score 774 · Answer 5 · 2017-10-07 04:10:12

774

Views

Details of why measurable functions of independent random variables independent.

Published on 07 Oct 2017 - 4:10

score 37 · Answer 6 · 2017-10-15 01:13:06

37

Views

$f_n$ measureable $\implies$ $m\left(\bigcap_{k=0}^\infty F_k\right)=0$

Published on 15 Oct 2017 - 1:13

score 330 · Answer 7 · 2026-03-29 17:31:43

330

Views

If $\int_{ \mathbb{R} } f^{2n} d \lambda = \int_{ \mathbb{R} } f d \lambda $ then $\lambda (f^{-1}((1,\infty)))=0$

Published on 29 Mar 2026 - 17:31

score 119 · Answer 8 · 2017-10-18 16:54:10

119

Views

Finding a Borel measurable function

Published on 18 Oct 2017 - 16:54

score 183 · Answer 9 · 2017-10-21 09:20:29

183

Views

Measurable function (Composite function)

Published on 21 Oct 2017 - 9:20

score 85 · Answer 10 · 2017-10-26 07:57:42

85

Views

measurability of functions; sub sigma field and completion

Published on 26 Oct 2017 - 7:57

score 43 · Answer 11 · 2017-11-02 03:01:11

43

Views

Show that $\mu(\{x:f(x)>0\}) > 0 \implies \mu(\{x:f(x)>1/N\}) > 0$

Published on 02 Nov 2017 - 3:01

score 18 · Answer 12 · 2017-11-05 13:17:33

18

Views

Finding conditions on X so T be continuous

Published on 05 Nov 2017 - 13:17

score 259 · Answer 13 · 2017-11-07 20:35:14

259

Views

Real analysis: measurable function $f$ and non measurable function $g$ such that their composition is measurable

Published on 07 Nov 2017 - 20:35

score 244 · Answer 14 · 2017-11-08 16:13:21

244

Views

Does there exist a non-measurable function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f ^{−1} (y) $is measurable for any $y \in \mathbb{R}$?

Published on 08 Nov 2017 - 16:13

score 17 · Answer 15 · 2026-03-12 17:58:29

17

Views

Measurability of $[f<g]$ where $f$, $g$ are $\mathbb{R}\cup\{\pm\infty\}$-valued

Published on 12 Mar 2026 - 17:58

List Question

Suppose $f$ is a measurable function and $f(x) > 0$ for all $x$. Let $g(x) = \frac{1}{f(x)}$. Prove that g is a measurable function.

Suppose $(X, \mathcal{A})$ is a measurable space, $f$ is a real valued function, and $\{x : f(x) > r\} \in \mathcal{A} $ for each rational number $r$.

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue measurable, prove that there exists a Borel measurable function $g$ s.t. $f = g$ a.e.

Embedding of some function spaces

Details of why measurable functions of independent random variables independent.

$f_n$ measureable $\implies$ $m\left(\bigcap_{k=0}^\infty F_k\right)=0$

If $\int_{ \mathbb{R} } f^{2n} d \lambda = \int_{ \mathbb{R} } f d \lambda $ then $\lambda (f^{-1}((1,\infty)))=0$

Finding a Borel measurable function

Measurable function (Composite function)

measurability of functions; sub sigma field and completion

Show that $\mu(\{x:f(x)>0\}) > 0 \implies \mu(\{x:f(x)>1/N\}) > 0$

Finding conditions on X so T be continuous

Real analysis: measurable function $f$ and non measurable function $g$ such that their composition is measurable

Does there exist a non-measurable function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f ^{−1} (y) $is measurable for any $y \in \mathbb{R}$?

Measurability of $[f<g]$ where $f$, $g$ are $\mathbb{R}\cup\{\pm\infty\}$-valued

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