Why $int^infty_01_>alphadalpha=|f(x)|$?Show that $int|f(x)|dx=int_0^infty m(E_alpha)dalpha$$e^-x^alpha$ is Lebesgue-integrable on $[0,infty)$ for $alpha>0$Show that the characteristic function of $mathbbQ$ is Lebesgue integrable.Lebesgue integral of f over R is bigger or equal to alpha times measure of $E_alpha?$$int_mathbbR^n frac1(lVert x-y rVert)^n- alpha dmu (y) = (n-alpha)int _[0,+infty[ r^alpha - n -1 mu(B(x,r)) dlambda (r)$For what $alpha > 0$ do we have $int_D_alpha f dlambda < + infty$ ?Show that $intlimits_[a,b]fdlambda = intlimits_(a,b) fdlambda$, for $-infty<a,b<infty$.Show that a differentiable function $f$, where $f(x)$ and $f'(x)$ are integrable, has a value of $0$ when Lebesgue integrated over the real numbers.DMCT $int fdmu=lim_ntoinfty int f_n dmu$ equivalent to $lim_ntoinftyint mid f_n -f mid dmu=0$?Show that $alpha f + beta g$ is measurable when $f,g$ are measurable.
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Why $int^infty_01_dalpha=|f(x)|$?
Show that $int|f(x)|dx=int_0^infty m(E_alpha)dalpha$$e^-x^alpha$ is Lebesgue-integrable on $[0,infty)$ for $alpha>0$Show that the characteristic function of $mathbbQ$ is Lebesgue integrable.Lebesgue integral of f over R is bigger or equal to alpha times measure of $E_alpha?$$int_mathbbR^n frac1(lVert x-y rVert)^n- alpha dmu (y) = (n-alpha)int _[0,+infty[ r^alpha - n -1 mu(B(x,r)) dlambda (r)$For what $alpha > 0$ do we have $int_D_alpha f dlambda < + infty$ ?Show that $intlimits_[a,b]fdlambda = intlimits_(a,b) fdlambda$, for $-infty<a,b<infty$.Show that a differentiable function $f$, where $f(x)$ and $f'(x)$ are integrable, has a value of $0$ when Lebesgue integrated over the real numbers.DMCT $int fdmu=lim_ntoinfty int f_n dmu$ equivalent to $lim_ntoinftyint mid f_n -f mid dmu=0$?Show that $alpha f + beta g$ is measurable when $f,g$ are measurable.
$begingroup$
$f$ is integrable on $mathbbR^d$. For each $alpha>0$, let $E_alpha=>alpha$. Show that
$$int_mathbbR^d|f(x)|dx=int^infty_0m(E_alpha)dalpha$$
I saw a proof of the above that uses the fact that
$$int^infty_01_dalpha=|f(x)|$$
Why is this true? Intuitively it seems like we're "stacking" the values of the characteristic function until we get to the value of $f$.
Any help is much appreciated
measure-theory proof-explanation lebesgue-integral
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
add a comment |
$begingroup$
$f$ is integrable on $mathbbR^d$. For each $alpha>0$, let $E_alpha=>alpha$. Show that
$$int_mathbbR^d|f(x)|dx=int^infty_0m(E_alpha)dalpha$$
I saw a proof of the above that uses the fact that
$$int^infty_01_dalpha=|f(x)|$$
Why is this true? Intuitively it seems like we're "stacking" the values of the characteristic function until we get to the value of $f$.
Any help is much appreciated
measure-theory proof-explanation lebesgue-integral
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
add a comment |
$begingroup$
$f$ is integrable on $mathbbR^d$. For each $alpha>0$, let $E_alpha=>alpha$. Show that
$$int_mathbbR^d|f(x)|dx=int^infty_0m(E_alpha)dalpha$$
I saw a proof of the above that uses the fact that
$$int^infty_01_dalpha=|f(x)|$$
Why is this true? Intuitively it seems like we're "stacking" the values of the characteristic function until we get to the value of $f$.
Any help is much appreciated
measure-theory proof-explanation lebesgue-integral
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
$f$ is integrable on $mathbbR^d$. For each $alpha>0$, let $E_alpha=>alpha$. Show that
$$int_mathbbR^d|f(x)|dx=int^infty_0m(E_alpha)dalpha$$
I saw a proof of the above that uses the fact that
$$int^infty_01_dalpha=|f(x)|$$
Why is this true? Intuitively it seems like we're "stacking" the values of the characteristic function until we get to the value of $f$.
Any help is much appreciated
measure-theory proof-explanation lebesgue-integral
measure-theory proof-explanation lebesgue-integral
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
asked Mar 13 at 10:07
Joe Man AnalysisJoe Man Analysis
56419
56419
add a comment |
add a comment |
1 Answer
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$begingroup$
$1_>alpha $ is nothing but the charateristic function of the interval $(0,|f(x)|)$ so its integral equals the measure of this interval.
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
add a comment |
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$begingroup$
$1_>alpha $ is nothing but the charateristic function of the interval $(0,|f(x)|)$ so its integral equals the measure of this interval.
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
add a comment |
$begingroup$
$1_>alpha $ is nothing but the charateristic function of the interval $(0,|f(x)|)$ so its integral equals the measure of this interval.
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
add a comment |
$begingroup$
$1_>alpha $ is nothing but the charateristic function of the interval $(0,|f(x)|)$ so its integral equals the measure of this interval.
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
$endgroup$
$1_>alpha $ is nothing but the charateristic function of the interval $(0,|f(x)|)$ so its integral equals the measure of this interval.
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
answered Mar 13 at 10:10
Kavi Rama MurthyKavi Rama Murthy
68.1k53068
68.1k53068
add a comment |
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