About two functions whose Lebesgue integral on all sets of a $sigma-$algebra are equal
$begingroup$
Let $X$ be an infinite set and $mathcal{F}$ be a $sigma-$algrbra with infinite sets on $X$. Given on $X$ a measure $mu$.
Let $f$ and $g$ be two $mathcal{F}-$measurable functions. Is it necessary that if $$ int_A f dmu = int_A g dmu, forall A in mathcal{F}$$
then $f = g$ ($mu-$a.e)?
I feel that the answer for this is positive, but I can't prove the statement above nor give a counter-example. Please give me a hint. Thank you.
measure-theory
asked Jan 14 at 4:59
ElementXElementX
389111
$endgroup$
add a comment |
$begingroup$
Let $X$ be an infinite set and $mathcal{F}$ be a $sigma-$algrbra with infinite sets on $X$. Given on $X$ a measure $mu$.
Let $f$ and $g$ be two $mathcal{F}-$measurable functions. Is it necessary that if $$ int_A f dmu = int_A g dmu, forall A in mathcal{F}$$
then $f = g$ ($mu-$a.e)?
I feel that the answer for this is positive, but I can't prove the statement above nor give a counter-example. Please give me a hint. Thank you.
measure-theory
asked Jan 14 at 4:59
ElementXElementX
389111
$endgroup$
1
$begingroup$
I'm assuming that you mean both $f$ and $g$ are $mathcal{F}$-measurable?
$endgroup$
– Keen-ameteur
Jan 14 at 5:32
$begingroup$
Yeah... I forgot to state it clearly. I will edit
$endgroup$
– ElementX
Jan 14 at 6:29
add a comment |
$begingroup$
Let $X$ be an infinite set and $mathcal{F}$ be a $sigma-$algrbra with infinite sets on $X$. Given on $X$ a measure $mu$.
Let $f$ and $g$ be two $mathcal{F}-$measurable functions. Is it necessary that if $$ int_A f dmu = int_A g dmu, forall A in mathcal{F}$$
then $f = g$ ($mu-$a.e)?
I feel that the answer for this is positive, but I can't prove the statement above nor give a counter-example. Please give me a hint. Thank you.
measure-theory
asked Jan 14 at 4:59
ElementXElementX
389111
$endgroup$
Let $X$ be an infinite set and $mathcal{F}$ be a $sigma-$algrbra with infinite sets on $X$. Given on $X$ a measure $mu$.
Let $f$ and $g$ be two $mathcal{F}-$measurable functions. Is it necessary that if $$ int_A f dmu = int_A g dmu, forall A in mathcal{F}$$
then $f = g$ ($mu-$a.e)?
I feel that the answer for this is positive, but I can't prove the statement above nor give a counter-example. Please give me a hint. Thank you.
measure-theory
measure-theory
asked Jan 14 at 4:59
ElementXElementX
389111
asked Jan 14 at 4:59
ElementXElementX
389111
edited Jan 14 at 6:30
ElementX
asked Jan 14 at 4:59
ElementXElementX
389111
asked Jan 14 at 4:59
ElementXElementX
389111
asked Jan 14 at 4:59
ElementXElementX
389111
389111
1
$begingroup$
I'm assuming that you mean both $f$ and $g$ are $mathcal{F}$-measurable?
$endgroup$
– Keen-ameteur
Jan 14 at 5:32
$begingroup$
Yeah... I forgot to state it clearly. I will edit
$endgroup$
– ElementX
Jan 14 at 6:29
add a comment |
1
$begingroup$
I'm assuming that you mean both $f$ and $g$ are $mathcal{F}$-measurable?
$endgroup$
– Keen-ameteur
Jan 14 at 5:32
$begingroup$
Yeah... I forgot to state it clearly. I will edit
$endgroup$
– ElementX
Jan 14 at 6:29
1
1
$begingroup$
I'm assuming that you mean both $f$ and $g$ are $mathcal{F}$-measurable?
$endgroup$
– Keen-ameteur
Jan 14 at 5:32
$begingroup$
I'm assuming that you mean both $f$ and $g$ are $mathcal{F}$-measurable?
$endgroup$
– Keen-ameteur
Jan 14 at 5:32
$begingroup$
Yeah... I forgot to state it clearly. I will edit
$endgroup$
– ElementX
Jan 14 at 6:29
$begingroup$
Yeah... I forgot to state it clearly. I will edit
$endgroup$
– ElementX
Jan 14 at 6:29
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes. Assume $f$ and $g$ are $mathcal{F}$-measurable. Then by assumption you have
$$0=int_{{f>g}}fdmu-int_{{f>g}}gdmu=int_{{f>g}}(f-g)dmu$$
Since $(f-g)cdot1_{{f>g}}ge 0$, you have $mu({f>g})=0$ and $mu({f<g})=0$ analogously.
Remark: This property is used to define conditional expectation, which is an important concept in stochastics.
answered Jan 14 at 5:45
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Hint:
${ fneq g}= underset{nin mathbb{N}}{bigcup} { f-g geq frac{1}{n} } cup underset{nin mathbb{N}}{bigcup} { g-f geq frac{1}{n} } $
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Yes. Assume $f$ and $g$ are $mathcal{F}$-measurable. Then by assumption you have
$$0=int_{{f>g}}fdmu-int_{{f>g}}gdmu=int_{{f>g}}(f-g)dmu$$
Since $(f-g)cdot1_{{f>g}}ge 0$, you have $mu({f>g})=0$ and $mu({f<g})=0$ analogously.
Remark: This property is used to define conditional expectation, which is an important concept in stochastics.
answered Jan 14 at 5:45
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Yes. Assume $f$ and $g$ are $mathcal{F}$-measurable. Then by assumption you have
$$0=int_{{f>g}}fdmu-int_{{f>g}}gdmu=int_{{f>g}}(f-g)dmu$$
Since $(f-g)cdot1_{{f>g}}ge 0$, you have $mu({f>g})=0$ and $mu({f<g})=0$ analogously.
Remark: This property is used to define conditional expectation, which is an important concept in stochastics.
answered Jan 14 at 5:45
user408858user408858
482213
$endgroup$
add a comment |
$begingroup$
Yes. Assume $f$ and $g$ are $mathcal{F}$-measurable. Then by assumption you have
$$0=int_{{f>g}}fdmu-int_{{f>g}}gdmu=int_{{f>g}}(f-g)dmu$$
Since $(f-g)cdot1_{{f>g}}ge 0$, you have $mu({f>g})=0$ and $mu({f<g})=0$ analogously.
Remark: This property is used to define conditional expectation, which is an important concept in stochastics.
answered Jan 14 at 5:45
user408858user408858
482213
$endgroup$
Yes. Assume $f$ and $g$ are $mathcal{F}$-measurable. Then by assumption you have
$$0=int_{{f>g}}fdmu-int_{{f>g}}gdmu=int_{{f>g}}(f-g)dmu$$
Since $(f-g)cdot1_{{f>g}}ge 0$, you have $mu({f>g})=0$ and $mu({f<g})=0$ analogously.
Remark: This property is used to define conditional expectation, which is an important concept in stochastics.
answered Jan 14 at 5:45
user408858user408858
482213
edited Jan 14 at 16:19
answered Jan 14 at 5:45
user408858user408858
482213
answered Jan 14 at 5:45
user408858user408858
482213
answered Jan 14 at 5:45
user408858user408858
482213
482213
add a comment |
add a comment |
$begingroup$
Hint:
${ fneq g}= underset{nin mathbb{N}}{bigcup} { f-g geq frac{1}{n} } cup underset{nin mathbb{N}}{bigcup} { g-f geq frac{1}{n} } $
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
$endgroup$
add a comment |
$begingroup$
Hint:
${ fneq g}= underset{nin mathbb{N}}{bigcup} { f-g geq frac{1}{n} } cup underset{nin mathbb{N}}{bigcup} { g-f geq frac{1}{n} } $
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
$endgroup$
add a comment |
$begingroup$
Hint:
${ fneq g}= underset{nin mathbb{N}}{bigcup} { f-g geq frac{1}{n} } cup underset{nin mathbb{N}}{bigcup} { g-f geq frac{1}{n} } $
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
$endgroup$
Hint:
${ fneq g}= underset{nin mathbb{N}}{bigcup} { f-g geq frac{1}{n} } cup underset{nin mathbb{N}}{bigcup} { g-f geq frac{1}{n} } $
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
answered Jan 14 at 5:42
Keen-ameteurKeen-ameteur
1,367316
1,367316
add a comment |
add a comment |
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1
$begingroup$
I'm assuming that you mean both $f$ and $g$ are $mathcal{F}$-measurable?
$endgroup$
– Keen-ameteur
Jan 14 at 5:32
$begingroup$
Yeah... I forgot to state it clearly. I will edit
$endgroup$
– ElementX
Jan 14 at 6:29