Question

Why is not $\infty$ allowed as a values of Lebesgue integral?

score 42 · Answer 1 · 2018-11-12 06:16:46

42

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Why is not $\infty$ allowed as a values of Lebesgue integral?

Published on 12 Nov 2018 - 6:16

score 471 · Answer 2 · 2026-03-25 17:40:39

471

Views

Is it possible to prove Fubini’s Theorem without Dynkin’s Theorem or the Monotone Class Theorem?

Published on 25 Mar 2026 - 17:40

score 97 · Answer 3 · 2026-03-25 09:09:17

97

Views

Can you prove Lusin's theorem without using approximation of $L_1$ functions?

Published on 25 Mar 2026 - 9:09

score 45 · Answer 4 · 2018-11-13 19:08:43

45

Views

Convergence of sequences of periodic functions

Published on 13 Nov 2018 - 19:08

score 224 · Answer 5 · 2018-11-13 22:26:45

224

Views

Prove that if $f^2(x)$ is Lebesgue integrable on E, so is $f(x)$.

Published on 13 Nov 2018 - 22:26

score 66 · Answer 6 · 2018-11-14 05:51:08

66

Views

How to prove this result somewhat similar to Du Bois-Reymond's Lemma?

Published on 14 Nov 2018 - 5:51

score 106 · Answer 7 · 2026-02-23 04:19:04

106

Views

construction of step functions to show integrability using Beppo Levi's theorem

Published on 23 Feb 2026 - 4:19

score 465 · Answer 8 · 2018-11-14 11:44:56

465

Views

Convergence of Lebesgue integral of an indicator function

Published on 14 Nov 2018 - 11:44

score 117 · Answer 9 · 2018-11-14 18:09:29

117

Views

Convergence of a sequence of measures of sets implies convergence of the indicator functions.

Published on 14 Nov 2018 - 18:09

score 560 · Answer 10 · 2026-03-25 03:24:09

560

Views

Generalize Jensen's Integral Inequality to the product of two functions

Published on 25 Mar 2026 - 3:24

score 215 · Answer 11 · 2026-03-25 03:21:16

215

Views

Counterexample of the converse of Jensen's inequality

Published on 25 Mar 2026 - 3:21

score 207 · Answer 12 · 2018-11-15 04:24:47

207

Views

Prove that if $f$ is nonnegative, measurable and $E_k \nearrow E$, then $\lim_k \int_{E_k} f = \int_E f$.

Published on 15 Nov 2018 - 4:24

score 45 · Answer 13 · 2018-11-15 19:34:32

45

Views

$\int|f|^\alpha|g|^{1-\alpha}d\mu\le(\int|f|d\mu)^\alpha(\int|g|d\mu)^{1-\alpha}$

Published on 15 Nov 2018 - 19:34

score 283 · Answer 14 · 2018-11-15 22:50:28

283

Views

In a measurable partition of an interval, the sum of the measures of the subsets in the partition equals the length of the interval

Published on 15 Nov 2018 - 22:50

score 319 · Answer 15 · 2018-11-16 23:25:56

319

Views

Hardy Littlewood maximal function $|h(x)|<|h^*(x)|$ for almost every x

Published on 16 Nov 2018 - 23:25

List Question

Why is not $\infty$ allowed as a values of Lebesgue integral?

Is it possible to prove Fubini’s Theorem without Dynkin’s Theorem or the Monotone Class Theorem?

Can you prove Lusin's theorem without using approximation of $L_1$ functions?

Convergence of sequences of periodic functions

Prove that if $f^2(x)$ is Lebesgue integrable on E, so is $f(x)$.

How to prove this result somewhat similar to Du Bois-Reymond's Lemma?

construction of step functions to show integrability using Beppo Levi's theorem

Convergence of Lebesgue integral of an indicator function

Convergence of a sequence of measures of sets implies convergence of the indicator functions.

Generalize Jensen's Integral Inequality to the product of two functions

Counterexample of the converse of Jensen's inequality

Prove that if $f$ is nonnegative, measurable and $E_k \nearrow E$, then $\lim_k \int_{E_k} f = \int_E f$.

$\int|f|^\alpha|g|^{1-\alpha}d\mu\le(\int|f|d\mu)^\alpha(\int|g|d\mu)^{1-\alpha}$

In a measurable partition of an interval, the sum of the measures of the subsets in the partition equals the length of the interval

Hardy Littlewood maximal function $|h(x)|<|h^*(x)|$ for almost every x

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