A small doubt about the dominated convergence theorem The Next CEO of Stack OverflowIs...
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A small doubt about the dominated convergence theorem
The Next CEO of Stack OverflowIs Lebesgue's Dominated Convergence Theorem a logical equivalence?Generalisation of Dominated Convergence TheoremLebesgue Convergence using The General Lebesgue Dominated Convergence TheoremVariant of dominated convergence theoremExample about Dominated Convergence TheoremDominated Convergence TheoremHypothesis of dominated convergence theoremBartle's proof of Lebesgue Dominated Convergence TheoremAn counterexample for the monotone convergence theorem and dominated convergence theoremTheorem similar to dominated convergence theorem
$begingroup$
Theorem $mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$
I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?
measure-theory convergence lebesgue-integral
$endgroup$
add a comment |
$begingroup$
Theorem $mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$
I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?
measure-theory convergence lebesgue-integral
$endgroup$
add a comment |
$begingroup$
Theorem $mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$
I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?
measure-theory convergence lebesgue-integral
$endgroup$
Theorem $mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X to mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X to mathbb R$ such that $|f_n(x)| leq |g(x)|$ for $mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $mu$-almost every point to some function $f : X to mathbb R$. Then $f$ is integrable and satisfies $$lim_n int f_n , dmu = int f , dmu.$$
I wanted to know if in the hypothesis $|f_n(x)| leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?
measure-theory convergence lebesgue-integral
measure-theory convergence lebesgue-integral
edited 35 mins ago
Rócherz
3,0013821
3,0013821
asked 47 mins ago
Ricardo FreireRicardo Freire
574211
574211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
$$
f_n(x) := frac{1}{n} mathbf{1}_{[0,n]}(x).
$$
Clearly, $f_n in L^1(mathbb{R})$ for each $n in mathbb{N}$. Moreover, $f_n to 0$ as $n to infty$ for each $x in mathbb{R}$. However,
begin{align*}
lim_{n to infty} int_{mathbb{R}} f_n,mathrm{d}m = lim_{n to infty} int_0^n frac{1}{n},mathrm{d}x = 1 neq 0.
end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrak{M},mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
$$
lim_{n to infty} int_E f_n,mathrm{d}mu = int_E f,mathrm{d}mu.
$$
In fact, one has $f_n to f$ strongly in $L^1(E)$.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.
$endgroup$
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_{[n,n+1]}$ on $mathbf R_{ge 0}$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_{ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have
$$
lim_{ntoinfty} int f_n = int lim_{ntoinfty}f_n
$$
since in this case, the left-hand side is $1$, but the right-hand side is $0$.
To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_{ge 0}$.
$endgroup$
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
$$
f_n(x) := frac{1}{n} mathbf{1}_{[0,n]}(x).
$$
Clearly, $f_n in L^1(mathbb{R})$ for each $n in mathbb{N}$. Moreover, $f_n to 0$ as $n to infty$ for each $x in mathbb{R}$. However,
begin{align*}
lim_{n to infty} int_{mathbb{R}} f_n,mathrm{d}m = lim_{n to infty} int_0^n frac{1}{n},mathrm{d}x = 1 neq 0.
end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrak{M},mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
$$
lim_{n to infty} int_E f_n,mathrm{d}mu = int_E f,mathrm{d}mu.
$$
In fact, one has $f_n to f$ strongly in $L^1(E)$.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.
$endgroup$
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
$$
f_n(x) := frac{1}{n} mathbf{1}_{[0,n]}(x).
$$
Clearly, $f_n in L^1(mathbb{R})$ for each $n in mathbb{N}$. Moreover, $f_n to 0$ as $n to infty$ for each $x in mathbb{R}$. However,
begin{align*}
lim_{n to infty} int_{mathbb{R}} f_n,mathrm{d}m = lim_{n to infty} int_0^n frac{1}{n},mathrm{d}x = 1 neq 0.
end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrak{M},mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
$$
lim_{n to infty} int_E f_n,mathrm{d}mu = int_E f,mathrm{d}mu.
$$
In fact, one has $f_n to f$ strongly in $L^1(E)$.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.
$endgroup$
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
$$
f_n(x) := frac{1}{n} mathbf{1}_{[0,n]}(x).
$$
Clearly, $f_n in L^1(mathbb{R})$ for each $n in mathbb{N}$. Moreover, $f_n to 0$ as $n to infty$ for each $x in mathbb{R}$. However,
begin{align*}
lim_{n to infty} int_{mathbb{R}} f_n,mathrm{d}m = lim_{n to infty} int_0^n frac{1}{n},mathrm{d}x = 1 neq 0.
end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrak{M},mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
$$
lim_{n to infty} int_E f_n,mathrm{d}mu = int_E f,mathrm{d}mu.
$$
In fact, one has $f_n to f$ strongly in $L^1(E)$.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.
$endgroup$
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence
$$
f_n(x) := frac{1}{n} mathbf{1}_{[0,n]}(x).
$$
Clearly, $f_n in L^1(mathbb{R})$ for each $n in mathbb{N}$. Moreover, $f_n to 0$ as $n to infty$ for each $x in mathbb{R}$. However,
begin{align*}
lim_{n to infty} int_{mathbb{R}} f_n,mathrm{d}m = lim_{n to infty} int_0^n frac{1}{n},mathrm{d}x = 1 neq 0.
end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
Theorem. Let $(f_n)$ be a sequence of measurable functions on a measure space $(X,mathfrak{M},mu)$ converging almost everywhere to a measurable function $f$. If $E subset X$ has finite measure and $(f_n)$ is bounded in $L^p(E)$ for some $1 < p < infty$, then
$$
lim_{n to infty} int_E f_n,mathrm{d}mu = int_E f,mathrm{d}mu.
$$
In fact, one has $f_n to f$ strongly in $L^1(E)$.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.
edited 23 mins ago
answered 30 mins ago
rolandcyprolandcyp
1,856315
1,856315
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_{[n,n+1]}$ on $mathbf R_{ge 0}$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_{ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have
$$
lim_{ntoinfty} int f_n = int lim_{ntoinfty}f_n
$$
since in this case, the left-hand side is $1$, but the right-hand side is $0$.
To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_{ge 0}$.
$endgroup$
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_{[n,n+1]}$ on $mathbf R_{ge 0}$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_{ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have
$$
lim_{ntoinfty} int f_n = int lim_{ntoinfty}f_n
$$
since in this case, the left-hand side is $1$, but the right-hand side is $0$.
To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_{ge 0}$.
$endgroup$
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_{[n,n+1]}$ on $mathbf R_{ge 0}$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_{ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have
$$
lim_{ntoinfty} int f_n = int lim_{ntoinfty}f_n
$$
since in this case, the left-hand side is $1$, but the right-hand side is $0$.
To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_{ge 0}$.
$endgroup$
In general, it is not sufficient that each $f_n$ be integrable without a dominating function. For instance, the functions $f_n = chi_{[n,n+1]}$ on $mathbf R_{ge 0}$ are all integrable, and $f_n(x) to 0$ for all $xin mathbf R_{ge 0}$, but they are not dominated by an integrable function $g$, and indeed we do not have
$$
lim_{ntoinfty} int f_n = int lim_{ntoinfty}f_n
$$
since in this case, the left-hand side is $1$, but the right-hand side is $0$.
To see why there is no dominating function $g$, such a function would have the property that $g(x)ge 1$ for each $xge 0$, so it would not be integrable on $mathbf R_{ge 0}$.
answered 29 mins ago
Alex OrtizAlex Ortiz
11.2k21441
11.2k21441
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
$begingroup$
I understood. Thanks a lot for the help
$endgroup$
– Ricardo Freire
16 mins ago
add a comment |
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