$int_A f = int_A g implies f = g$ a.e.












4














Consider a measure space $(Omega, mathcal{F}, mu)$ and let $f,g: Omega to mathbb{R}$ be $mathcal{F}$-measurable integrable functions. If



$$int_A f d mu = int_A g d mu$$



for all $A in mathcal{F}$, then $f = g$ $mu$-a.e.



Here is the proof my course notes provide:



It is sufficient to prove that $mu{f < g} = 0 = mu{g < f}$. By symmetry, we only prove the first equality.



Put $$h:= (g-f)I_{{g > f}}$$



Then $$int_Omega h dmu = int_{{f<g}}(g-f) = 0$$ since $f,g$ are integrable functions. Since $h geq 0$, it follows that $h = 0$ a.e. and thus $mu{g > f}
leq mu{h neq 0} = 0$. This ends the proof.




My question: Can we generalise this to integrable functions $f,g:
Omega to [-infty, infty]$ (note the change of codomain).




I think the proof generalises: the function $h$ is well defined because we can only have $infty - (-infty) = infty$. Is this correct?






measure-theory proof-verification lebesgue-integral almost-everywhere







share|cite|improve this question




















  • 1




    How do you define integration over this kind of functions? In particular what is the integral of $infty cdot chi_{{0}}$?
    – 0x539
    Jan 3 at 22:01












  • $1_0$? Do you mean $I_emptyset?$
    – Math_QED
    Jan 3 at 22:02






  • 1




    @Math_QED I presume that they mean the characteristic function of the singleton set containing $0$. In other words, what is $$int_{Omega} f(x),mathrm{d}mu $$ when $f$ is the function $$f(x) = begin{cases} infty, & x= 0 \ 0, & xne 0. end{cases} $$
    – Xander Henderson
    Jan 3 at 22:03








  • 2




    I define $int_Omega alpha I_A d mu := alpha mu(A)$ where $0. infty = 0 = infty. 0$. So, assuming lebesgue measure (and singeltons have measure $0$), that integral would be equal to $0$.
    – Math_QED
    Jan 3 at 22:05






  • 1




    Yes, if we restrict to $f<g$, and both $f(x),g(x)$ are infinite, we must have $f(x)=-infty, g(x)=+infty$, so $h$ indeed seems well defined.
    – Berci
    Jan 3 at 22:29
















4














Consider a measure space $(Omega, mathcal{F}, mu)$ and let $f,g: Omega to mathbb{R}$ be $mathcal{F}$-measurable integrable functions. If



$$int_A f d mu = int_A g d mu$$



for all $A in mathcal{F}$, then $f = g$ $mu$-a.e.



Here is the proof my course notes provide:



It is sufficient to prove that $mu{f < g} = 0 = mu{g < f}$. By symmetry, we only prove the first equality.



Put $$h:= (g-f)I_{{g > f}}$$



Then $$int_Omega h dmu = int_{{f<g}}(g-f) = 0$$ since $f,g$ are integrable functions. Since $h geq 0$, it follows that $h = 0$ a.e. and thus $mu{g > f}
leq mu{h neq 0} = 0$. This ends the proof.




My question: Can we generalise this to integrable functions $f,g:
Omega to [-infty, infty]$ (note the change of codomain).




I think the proof generalises: the function $h$ is well defined because we can only have $infty - (-infty) = infty$. Is this correct?






measure-theory proof-verification lebesgue-integral almost-everywhere







share|cite|improve this question




















  • 1




    How do you define integration over this kind of functions? In particular what is the integral of $infty cdot chi_{{0}}$?
    – 0x539
    Jan 3 at 22:01












  • $1_0$? Do you mean $I_emptyset?$
    – Math_QED
    Jan 3 at 22:02






  • 1




    @Math_QED I presume that they mean the characteristic function of the singleton set containing $0$. In other words, what is $$int_{Omega} f(x),mathrm{d}mu $$ when $f$ is the function $$f(x) = begin{cases} infty, & x= 0 \ 0, & xne 0. end{cases} $$
    – Xander Henderson
    Jan 3 at 22:03








  • 2




    I define $int_Omega alpha I_A d mu := alpha mu(A)$ where $0. infty = 0 = infty. 0$. So, assuming lebesgue measure (and singeltons have measure $0$), that integral would be equal to $0$.
    – Math_QED
    Jan 3 at 22:05






  • 1




    Yes, if we restrict to $f<g$, and both $f(x),g(x)$ are infinite, we must have $f(x)=-infty, g(x)=+infty$, so $h$ indeed seems well defined.
    – Berci
    Jan 3 at 22:29














4












4








4







Consider a measure space $(Omega, mathcal{F}, mu)$ and let $f,g: Omega to mathbb{R}$ be $mathcal{F}$-measurable integrable functions. If



$$int_A f d mu = int_A g d mu$$



for all $A in mathcal{F}$, then $f = g$ $mu$-a.e.



Here is the proof my course notes provide:



It is sufficient to prove that $mu{f < g} = 0 = mu{g < f}$. By symmetry, we only prove the first equality.



Put $$h:= (g-f)I_{{g > f}}$$



Then $$int_Omega h dmu = int_{{f<g}}(g-f) = 0$$ since $f,g$ are integrable functions. Since $h geq 0$, it follows that $h = 0$ a.e. and thus $mu{g > f}
leq mu{h neq 0} = 0$. This ends the proof.




My question: Can we generalise this to integrable functions $f,g:
Omega to [-infty, infty]$ (note the change of codomain).




I think the proof generalises: the function $h$ is well defined because we can only have $infty - (-infty) = infty$. Is this correct?






measure-theory proof-verification lebesgue-integral almost-everywhere







share|cite|improve this question















Consider a measure space $(Omega, mathcal{F}, mu)$ and let $f,g: Omega to mathbb{R}$ be $mathcal{F}$-measurable integrable functions. If



$$int_A f d mu = int_A g d mu$$



for all $A in mathcal{F}$, then $f = g$ $mu$-a.e.



Here is the proof my course notes provide:



It is sufficient to prove that $mu{f < g} = 0 = mu{g < f}$. By symmetry, we only prove the first equality.



Put $$h:= (g-f)I_{{g > f}}$$



Then $$int_Omega h dmu = int_{{f<g}}(g-f) = 0$$ since $f,g$ are integrable functions. Since $h geq 0$, it follows that $h = 0$ a.e. and thus $mu{g > f}
leq mu{h neq 0} = 0$. This ends the proof.




My question: Can we generalise this to integrable functions $f,g:
Omega to [-infty, infty]$ (note the change of codomain).




I think the proof generalises: the function $h$ is well defined because we can only have $infty - (-infty) = infty$. Is this correct?






measure-theory proof-verification lebesgue-integral almost-everywhere




measure-theory proof-verification lebesgue-integral almost-everywhere






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago

























asked Jan 3 at 21:56









Math_QED

7,18731449




7,18731449








  • 1




    How do you define integration over this kind of functions? In particular what is the integral of $infty cdot chi_{{0}}$?
    – 0x539
    Jan 3 at 22:01












  • $1_0$? Do you mean $I_emptyset?$
    – Math_QED
    Jan 3 at 22:02






  • 1




    @Math_QED I presume that they mean the characteristic function of the singleton set containing $0$. In other words, what is $$int_{Omega} f(x),mathrm{d}mu $$ when $f$ is the function $$f(x) = begin{cases} infty, & x= 0 \ 0, & xne 0. end{cases} $$
    – Xander Henderson
    Jan 3 at 22:03








  • 2




    I define $int_Omega alpha I_A d mu := alpha mu(A)$ where $0. infty = 0 = infty. 0$. So, assuming lebesgue measure (and singeltons have measure $0$), that integral would be equal to $0$.
    – Math_QED
    Jan 3 at 22:05






  • 1




    Yes, if we restrict to $f<g$, and both $f(x),g(x)$ are infinite, we must have $f(x)=-infty, g(x)=+infty$, so $h$ indeed seems well defined.
    – Berci
    Jan 3 at 22:29














  • 1




    How do you define integration over this kind of functions? In particular what is the integral of $infty cdot chi_{{0}}$?
    – 0x539
    Jan 3 at 22:01












  • $1_0$? Do you mean $I_emptyset?$
    – Math_QED
    Jan 3 at 22:02






  • 1




    @Math_QED I presume that they mean the characteristic function of the singleton set containing $0$. In other words, what is $$int_{Omega} f(x),mathrm{d}mu $$ when $f$ is the function $$f(x) = begin{cases} infty, & x= 0 \ 0, & xne 0. end{cases} $$
    – Xander Henderson
    Jan 3 at 22:03








  • 2




    I define $int_Omega alpha I_A d mu := alpha mu(A)$ where $0. infty = 0 = infty. 0$. So, assuming lebesgue measure (and singeltons have measure $0$), that integral would be equal to $0$.
    – Math_QED
    Jan 3 at 22:05






  • 1




    Yes, if we restrict to $f<g$, and both $f(x),g(x)$ are infinite, we must have $f(x)=-infty, g(x)=+infty$, so $h$ indeed seems well defined.
    – Berci
    Jan 3 at 22:29








1




1




How do you define integration over this kind of functions? In particular what is the integral of $infty cdot chi_{{0}}$?
– 0x539
Jan 3 at 22:01






How do you define integration over this kind of functions? In particular what is the integral of $infty cdot chi_{{0}}$?
– 0x539
Jan 3 at 22:01














$1_0$? Do you mean $I_emptyset?$
– Math_QED
Jan 3 at 22:02




$1_0$? Do you mean $I_emptyset?$
– Math_QED
Jan 3 at 22:02




1




1




@Math_QED I presume that they mean the characteristic function of the singleton set containing $0$. In other words, what is $$int_{Omega} f(x),mathrm{d}mu $$ when $f$ is the function $$f(x) = begin{cases} infty, & x= 0 \ 0, & xne 0. end{cases} $$
– Xander Henderson
Jan 3 at 22:03






@Math_QED I presume that they mean the characteristic function of the singleton set containing $0$. In other words, what is $$int_{Omega} f(x),mathrm{d}mu $$ when $f$ is the function $$f(x) = begin{cases} infty, & x= 0 \ 0, & xne 0. end{cases} $$
– Xander Henderson
Jan 3 at 22:03






2




2




I define $int_Omega alpha I_A d mu := alpha mu(A)$ where $0. infty = 0 = infty. 0$. So, assuming lebesgue measure (and singeltons have measure $0$), that integral would be equal to $0$.
– Math_QED
Jan 3 at 22:05




I define $int_Omega alpha I_A d mu := alpha mu(A)$ where $0. infty = 0 = infty. 0$. So, assuming lebesgue measure (and singeltons have measure $0$), that integral would be equal to $0$.
– Math_QED
Jan 3 at 22:05




1




1




Yes, if we restrict to $f<g$, and both $f(x),g(x)$ are infinite, we must have $f(x)=-infty, g(x)=+infty$, so $h$ indeed seems well defined.
– Berci
Jan 3 at 22:29




Yes, if we restrict to $f<g$, and both $f(x),g(x)$ are infinite, we must have $f(x)=-infty, g(x)=+infty$, so $h$ indeed seems well defined.
– Berci
Jan 3 at 22:29










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