Mixing
Functions are integrable with respect to a measure
Let $delta_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
Define $delta_x(E)=begin{cases}1,&text{if }x in Etext{ }\0,&text{if }xnotin Eend{cases}quad$
How can it be shown that every map $f: mathbb{R} to mathbb{R}$ is integrable with respect to $delta_x$?
I tried to use:
If $f: [a,b] to mathbb{R}$ is a Riemann integral, then $f in mathcal{L}(E,delta_x)$.
So: $int f delta_x= int_{a}^{b}f$
Here I don't know how to continue.
Or is there another way to prove that all $f: mathbb{R} to mathbb{R}$ are integrable with respect to $delta_x$?
measure-theory lebesgue-integral
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
add a comment |
Let $delta_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
Define $delta_x(E)=begin{cases}1,&text{if }x in Etext{ }\0,&text{if }xnotin Eend{cases}quad$
How can it be shown that every map $f: mathbb{R} to mathbb{R}$ is integrable with respect to $delta_x$?
I tried to use:
If $f: [a,b] to mathbb{R}$ is a Riemann integral, then $f in mathcal{L}(E,delta_x)$.
So: $int f delta_x= int_{a}^{b}f$
Here I don't know how to continue.
Or is there another way to prove that all $f: mathbb{R} to mathbb{R}$ are integrable with respect to $delta_x$?
measure-theory lebesgue-integral
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
"Since $delta_x$ seems to be a step function", no, it is not; it is a "measure".
– user587192
Nov 21 '18 at 18:57
Do you know the definition of $f$ being integrable with respect to $delta_x$? That should be the first step of doing this problem.
– user587192
Nov 21 '18 at 18:58
add a comment |
Let $delta_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
Define $delta_x(E)=begin{cases}1,&text{if }x in Etext{ }\0,&text{if }xnotin Eend{cases}quad$
How can it be shown that every map $f: mathbb{R} to mathbb{R}$ is integrable with respect to $delta_x$?
I tried to use:
If $f: [a,b] to mathbb{R}$ is a Riemann integral, then $f in mathcal{L}(E,delta_x)$.
So: $int f delta_x= int_{a}^{b}f$
Here I don't know how to continue.
Or is there another way to prove that all $f: mathbb{R} to mathbb{R}$ are integrable with respect to $delta_x$?
measure-theory lebesgue-integral
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
Let $delta_x$ be a measure on $mathcal{P}(mathbb{R^n})$.
Define $delta_x(E)=begin{cases}1,&text{if }x in Etext{ }\0,&text{if }xnotin Eend{cases}quad$
How can it be shown that every map $f: mathbb{R} to mathbb{R}$ is integrable with respect to $delta_x$?
I tried to use:
If $f: [a,b] to mathbb{R}$ is a Riemann integral, then $f in mathcal{L}(E,delta_x)$.
So: $int f delta_x= int_{a}^{b}f$
Here I don't know how to continue.
Or is there another way to prove that all $f: mathbb{R} to mathbb{R}$ are integrable with respect to $delta_x$?
measure-theory lebesgue-integral
measure-theory lebesgue-integral
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
edited Nov 21 '18 at 19:08
Olsgur
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
asked Nov 21 '18 at 18:40
OlsgurOlsgur
444
444
"Since $delta_x$ seems to be a step function", no, it is not; it is a "measure".
– user587192
Nov 21 '18 at 18:57
Do you know the definition of $f$ being integrable with respect to $delta_x$? That should be the first step of doing this problem.
– user587192
Nov 21 '18 at 18:58
add a comment |
"Since $delta_x$ seems to be a step function", no, it is not; it is a "measure".
– user587192
Nov 21 '18 at 18:57
Do you know the definition of $f$ being integrable with respect to $delta_x$? That should be the first step of doing this problem.
– user587192
Nov 21 '18 at 18:58
"Since $delta_x$ seems to be a step function", no, it is not; it is a "measure".
– user587192
Nov 21 '18 at 18:57
"Since $delta_x$ seems to be a step function", no, it is not; it is a "measure".
– user587192
Nov 21 '18 at 18:57
Do you know the definition of $f$ being integrable with respect to $delta_x$? That should be the first step of doing this problem.
– user587192
Nov 21 '18 at 18:58
Do you know the definition of $f$ being integrable with respect to $delta_x$? That should be the first step of doing this problem.
– user587192
Nov 21 '18 at 18:58
add a comment |
1 Answer
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One can directly check the definition of Lebesgue integrability. For any simple function $f$, one can directly verify that the integral is equal to $f(x)$. Then it extends to non-negative measurable $f$ by the definition of Lebesgue integral. If $f$ takes real values, then $int|f| ddelta_x= |f(x)|<infty$, so integrability holds.
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
One can directly check the definition of Lebesgue integrability. For any simple function $f$, one can directly verify that the integral is equal to $f(x)$. Then it extends to non-negative measurable $f$ by the definition of Lebesgue integral. If $f$ takes real values, then $int|f| ddelta_x= |f(x)|<infty$, so integrability holds.
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
add a comment |
One can directly check the definition of Lebesgue integrability. For any simple function $f$, one can directly verify that the integral is equal to $f(x)$. Then it extends to non-negative measurable $f$ by the definition of Lebesgue integral. If $f$ takes real values, then $int|f| ddelta_x= |f(x)|<infty$, so integrability holds.
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
add a comment |
One can directly check the definition of Lebesgue integrability. For any simple function $f$, one can directly verify that the integral is equal to $f(x)$. Then it extends to non-negative measurable $f$ by the definition of Lebesgue integral. If $f$ takes real values, then $int|f| ddelta_x= |f(x)|<infty$, so integrability holds.
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
One can directly check the definition of Lebesgue integrability. For any simple function $f$, one can directly verify that the integral is equal to $f(x)$. Then it extends to non-negative measurable $f$ by the definition of Lebesgue integral. If $f$ takes real values, then $int|f| ddelta_x= |f(x)|<infty$, so integrability holds.
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
answered Nov 21 '18 at 18:46
UchihaUchiha
9310
9310
add a comment |
add a comment |
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"Since $delta_x$ seems to be a step function", no, it is not; it is a "measure".
– user587192
Nov 21 '18 at 18:57
Do you know the definition of $f$ being integrable with respect to $delta_x$? That should be the first step of doing this problem.
– user587192
Nov 21 '18 at 18:58