Prove Lebesgue integrability for an (almost) trigonometric function
$begingroup$
Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$
As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?
As for integration, using the sigma-additive property of the integral, we can express the integral this way:
$intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$
Unfortunately, I don't exactly understand how to calculate this integral.
measure-theory lebesgue-integral lebesgue-measure
asked Jan 17 at 17:42
Don DraperDon Draper
539
$endgroup$
add a comment |
$begingroup$
Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$
As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?
As for integration, using the sigma-additive property of the integral, we can express the integral this way:
$intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$
Unfortunately, I don't exactly understand how to calculate this integral.
measure-theory lebesgue-integral lebesgue-measure
asked Jan 17 at 17:42
Don DraperDon Draper
539
$endgroup$
add a comment |
$begingroup$
Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$
As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?
As for integration, using the sigma-additive property of the integral, we can express the integral this way:
$intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$
Unfortunately, I don't exactly understand how to calculate this integral.
measure-theory lebesgue-integral lebesgue-measure
asked Jan 17 at 17:42
Don DraperDon Draper
539
$endgroup$
Let $X = [0,1]$, $mathfrak{M}$ - is a $sigma$ algebra of Lebesgue measurable subsets of $X$, $mu$ - Lebesgue measure on $mathfrak{M}$ Function $f:Xtomathbb{R} $ is defined as: $f(x) = sin{nx}$ if $x in (frac{1}{2^n},frac{1}{2^{n-1}}], n in mathbb{N}, f(0) = 0$
Prove that $f$ is Lebesgue integrable and calculate $intlimits_{X}fdmu$
As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct?
As for integration, using the sigma-additive property of the integral, we can express the integral this way:
$intlimits_{X}fdmu = sumlimits_{n=1}^{infty}intlimits_{A_n}fdmu, n in mathbb{N}$
Unfortunately, I don't exactly understand how to calculate this integral.
measure-theory lebesgue-integral lebesgue-measure
measure-theory lebesgue-integral lebesgue-measure
asked Jan 17 at 17:42
Don DraperDon Draper
539
asked Jan 17 at 17:42
Don DraperDon Draper
539
edited Jan 18 at 2:47
Don Draper
asked Jan 17 at 17:42
Don DraperDon Draper
539
asked Jan 17 at 17:42
Don DraperDon Draper
539
asked Jan 17 at 17:42
Don DraperDon Draper
539
539
add a comment |
add a comment |
1 Answer
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$begingroup$
The sequence
$$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and
$$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,
$$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$
answered Jan 18 at 5:50
RRLRRL
51k42573
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The sequence
$$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and
$$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,
$$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$
answered Jan 18 at 5:50
RRLRRL
51k42573
$endgroup$
add a comment |
$begingroup$
The sequence
$$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and
$$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,
$$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$
answered Jan 18 at 5:50
RRLRRL
51k42573
$endgroup$
add a comment |
$begingroup$
The sequence
$$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and
$$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,
$$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$
answered Jan 18 at 5:50
RRLRRL
51k42573
$endgroup$
The sequence
$$f_n(x) = sum_{k=1}^n sin kx , mathbf{1}_{left(2^{-k},2^{-k+1} right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < sin kx < 1$ for $x in left(2^{-k},2^{-k+1} right]$ we have $|f_n(x)| leqslant 1$ for all $x in [0,1]$. By the LDCT, $f$ is integrable and
$$int_{[0,1]} f , dmu= lim_{nto infty}int_{[0,1]} f_n , dmu = lim_{nto infty}sum_{k=1}^n int_{(2^{-k},2^{-k+1}]} sin kx , dmu \ = sum_{k=1}^infty int_{(2^{-k},2^{-k+1}]} sin kx , dmu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus,
$$int_{[0,1]} f , dmu = sum_{k=1}^infty int_{2^{-k}}^{2^{-k+1}} sin kx , dx \ = sum_{k=1}^infty frac{cos frac{k}{2^{k}} - cos frac{k}{2^{k-1}}}{k} \ underbrace{approx 0.605359}_{text{WolframAlpha}}$$
answered Jan 18 at 5:50
RRLRRL
51k42573
answered Jan 18 at 5:50
RRLRRL
51k42573
answered Jan 18 at 5:50
RRLRRL
51k42573
answered Jan 18 at 5:50
RRLRRL
51k42573
51k42573
add a comment |
add a comment |
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