Prove an equivalent condition to Lebesgue integrability












0












$begingroup$



Let $f$ be a non-negative, measurable, real-valued function defined on all of $mathbb{R}$. Show that $f$ is Lebesgue integrable on $mathbb{R}$ if and only if the series $sum_{k=0}^{infty}km(E_k)$ converges, where $E_k = {x|kle f(x)<k+1}$ and $m$ denotes Lebesgue measure.




The direction $(implies)$ is simply done by countable additivity of integration, since $E_k$'s are pairwise disjoint and we can write
$$infty>int_{mathbb{R}} f = sum_{kge0}int_{E_k}fge sum_{kge0}km(E_k). $$
Hence the convergence of the series.



But I was stuck on the opposite direction and need some help. Thanks!






real-analysis integration measure-theory lebesgue-integral







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What about the case $f(x) = frac{1}{x} chi_{[1, infty)}(x)$ in which case $m(E_0) = infty$, $m(E_1) = m(E_2) = cdots = 0$ but $f$ is not $L^1$?
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:36










  • $begingroup$
    @DanielSchepler Thanks for the comment. So you're saying that this is one example implying the series diverges, but $fnotin L^1$? But this is the same as saying that f is $L^1$ then the series converges, right? I guess you were trying to give a counter example...
    $endgroup$
    – Alex
    Jan 8 at 21:44












  • $begingroup$
    No, the series converges since the measure theory convention, especially in cases like this, is $0 cdot infty = 0$. (But on the other hand, if you count that as divergent, then $f(x) = frac{1}{x^2} chi_{[1, infty)}(x)$ would form a counterexample to the forward direction with the same $m(E_k)$.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:46










  • $begingroup$
    @DanielSchepler Oh, I see. So the statement is actually false. They are not equivalent at all. But if we take the convention $0cdotinfty=0$, then is the above proof for the forward direction correct?
    $endgroup$
    – Alex
    Jan 8 at 21:57






  • 1




    $begingroup$
    I think a correct statement could go something like: if $F_k = { x mid frac{1}{k+1} < f(x) le frac{1}{k} }$, then $f$ is Lebesgue integrable if and only if $sum_{k=1}^infty k m(E_k)$ and $sum_{k=1}^infty frac{1}{k} m(F_k)$ are both convergent. (In other words, there are two main concerns for integrability: that not too much of a function is "large", and that how much of a function is "small" increases sufficiently slowly.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 22:05
















0












$begingroup$



Let $f$ be a non-negative, measurable, real-valued function defined on all of $mathbb{R}$. Show that $f$ is Lebesgue integrable on $mathbb{R}$ if and only if the series $sum_{k=0}^{infty}km(E_k)$ converges, where $E_k = {x|kle f(x)<k+1}$ and $m$ denotes Lebesgue measure.




The direction $(implies)$ is simply done by countable additivity of integration, since $E_k$'s are pairwise disjoint and we can write
$$infty>int_{mathbb{R}} f = sum_{kge0}int_{E_k}fge sum_{kge0}km(E_k). $$
Hence the convergence of the series.



But I was stuck on the opposite direction and need some help. Thanks!






real-analysis integration measure-theory lebesgue-integral







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What about the case $f(x) = frac{1}{x} chi_{[1, infty)}(x)$ in which case $m(E_0) = infty$, $m(E_1) = m(E_2) = cdots = 0$ but $f$ is not $L^1$?
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:36










  • $begingroup$
    @DanielSchepler Thanks for the comment. So you're saying that this is one example implying the series diverges, but $fnotin L^1$? But this is the same as saying that f is $L^1$ then the series converges, right? I guess you were trying to give a counter example...
    $endgroup$
    – Alex
    Jan 8 at 21:44












  • $begingroup$
    No, the series converges since the measure theory convention, especially in cases like this, is $0 cdot infty = 0$. (But on the other hand, if you count that as divergent, then $f(x) = frac{1}{x^2} chi_{[1, infty)}(x)$ would form a counterexample to the forward direction with the same $m(E_k)$.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:46










  • $begingroup$
    @DanielSchepler Oh, I see. So the statement is actually false. They are not equivalent at all. But if we take the convention $0cdotinfty=0$, then is the above proof for the forward direction correct?
    $endgroup$
    – Alex
    Jan 8 at 21:57






  • 1




    $begingroup$
    I think a correct statement could go something like: if $F_k = { x mid frac{1}{k+1} < f(x) le frac{1}{k} }$, then $f$ is Lebesgue integrable if and only if $sum_{k=1}^infty k m(E_k)$ and $sum_{k=1}^infty frac{1}{k} m(F_k)$ are both convergent. (In other words, there are two main concerns for integrability: that not too much of a function is "large", and that how much of a function is "small" increases sufficiently slowly.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 22:05














0












0








0





$begingroup$



Let $f$ be a non-negative, measurable, real-valued function defined on all of $mathbb{R}$. Show that $f$ is Lebesgue integrable on $mathbb{R}$ if and only if the series $sum_{k=0}^{infty}km(E_k)$ converges, where $E_k = {x|kle f(x)<k+1}$ and $m$ denotes Lebesgue measure.




The direction $(implies)$ is simply done by countable additivity of integration, since $E_k$'s are pairwise disjoint and we can write
$$infty>int_{mathbb{R}} f = sum_{kge0}int_{E_k}fge sum_{kge0}km(E_k). $$
Hence the convergence of the series.



But I was stuck on the opposite direction and need some help. Thanks!






real-analysis integration measure-theory lebesgue-integral







share|cite|improve this question









$endgroup$





Let $f$ be a non-negative, measurable, real-valued function defined on all of $mathbb{R}$. Show that $f$ is Lebesgue integrable on $mathbb{R}$ if and only if the series $sum_{k=0}^{infty}km(E_k)$ converges, where $E_k = {x|kle f(x)<k+1}$ and $m$ denotes Lebesgue measure.




The direction $(implies)$ is simply done by countable additivity of integration, since $E_k$'s are pairwise disjoint and we can write
$$infty>int_{mathbb{R}} f = sum_{kge0}int_{E_k}fge sum_{kge0}km(E_k). $$
Hence the convergence of the series.



But I was stuck on the opposite direction and need some help. Thanks!






real-analysis integration measure-theory lebesgue-integral




real-analysis integration measure-theory lebesgue-integral






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 21:30









AlexAlex

757




757








  • 1




    $begingroup$
    What about the case $f(x) = frac{1}{x} chi_{[1, infty)}(x)$ in which case $m(E_0) = infty$, $m(E_1) = m(E_2) = cdots = 0$ but $f$ is not $L^1$?
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:36










  • $begingroup$
    @DanielSchepler Thanks for the comment. So you're saying that this is one example implying the series diverges, but $fnotin L^1$? But this is the same as saying that f is $L^1$ then the series converges, right? I guess you were trying to give a counter example...
    $endgroup$
    – Alex
    Jan 8 at 21:44












  • $begingroup$
    No, the series converges since the measure theory convention, especially in cases like this, is $0 cdot infty = 0$. (But on the other hand, if you count that as divergent, then $f(x) = frac{1}{x^2} chi_{[1, infty)}(x)$ would form a counterexample to the forward direction with the same $m(E_k)$.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:46










  • $begingroup$
    @DanielSchepler Oh, I see. So the statement is actually false. They are not equivalent at all. But if we take the convention $0cdotinfty=0$, then is the above proof for the forward direction correct?
    $endgroup$
    – Alex
    Jan 8 at 21:57






  • 1




    $begingroup$
    I think a correct statement could go something like: if $F_k = { x mid frac{1}{k+1} < f(x) le frac{1}{k} }$, then $f$ is Lebesgue integrable if and only if $sum_{k=1}^infty k m(E_k)$ and $sum_{k=1}^infty frac{1}{k} m(F_k)$ are both convergent. (In other words, there are two main concerns for integrability: that not too much of a function is "large", and that how much of a function is "small" increases sufficiently slowly.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 22:05














  • 1




    $begingroup$
    What about the case $f(x) = frac{1}{x} chi_{[1, infty)}(x)$ in which case $m(E_0) = infty$, $m(E_1) = m(E_2) = cdots = 0$ but $f$ is not $L^1$?
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:36










  • $begingroup$
    @DanielSchepler Thanks for the comment. So you're saying that this is one example implying the series diverges, but $fnotin L^1$? But this is the same as saying that f is $L^1$ then the series converges, right? I guess you were trying to give a counter example...
    $endgroup$
    – Alex
    Jan 8 at 21:44












  • $begingroup$
    No, the series converges since the measure theory convention, especially in cases like this, is $0 cdot infty = 0$. (But on the other hand, if you count that as divergent, then $f(x) = frac{1}{x^2} chi_{[1, infty)}(x)$ would form a counterexample to the forward direction with the same $m(E_k)$.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 21:46










  • $begingroup$
    @DanielSchepler Oh, I see. So the statement is actually false. They are not equivalent at all. But if we take the convention $0cdotinfty=0$, then is the above proof for the forward direction correct?
    $endgroup$
    – Alex
    Jan 8 at 21:57






  • 1




    $begingroup$
    I think a correct statement could go something like: if $F_k = { x mid frac{1}{k+1} < f(x) le frac{1}{k} }$, then $f$ is Lebesgue integrable if and only if $sum_{k=1}^infty k m(E_k)$ and $sum_{k=1}^infty frac{1}{k} m(F_k)$ are both convergent. (In other words, there are two main concerns for integrability: that not too much of a function is "large", and that how much of a function is "small" increases sufficiently slowly.)
    $endgroup$
    – Daniel Schepler
    Jan 8 at 22:05








1




1




$begingroup$
What about the case $f(x) = frac{1}{x} chi_{[1, infty)}(x)$ in which case $m(E_0) = infty$, $m(E_1) = m(E_2) = cdots = 0$ but $f$ is not $L^1$?
$endgroup$
– Daniel Schepler
Jan 8 at 21:36




$begingroup$
What about the case $f(x) = frac{1}{x} chi_{[1, infty)}(x)$ in which case $m(E_0) = infty$, $m(E_1) = m(E_2) = cdots = 0$ but $f$ is not $L^1$?
$endgroup$
– Daniel Schepler
Jan 8 at 21:36












$begingroup$
@DanielSchepler Thanks for the comment. So you're saying that this is one example implying the series diverges, but $fnotin L^1$? But this is the same as saying that f is $L^1$ then the series converges, right? I guess you were trying to give a counter example...
$endgroup$
– Alex
Jan 8 at 21:44






$begingroup$
@DanielSchepler Thanks for the comment. So you're saying that this is one example implying the series diverges, but $fnotin L^1$? But this is the same as saying that f is $L^1$ then the series converges, right? I guess you were trying to give a counter example...
$endgroup$
– Alex
Jan 8 at 21:44














$begingroup$
No, the series converges since the measure theory convention, especially in cases like this, is $0 cdot infty = 0$. (But on the other hand, if you count that as divergent, then $f(x) = frac{1}{x^2} chi_{[1, infty)}(x)$ would form a counterexample to the forward direction with the same $m(E_k)$.)
$endgroup$
– Daniel Schepler
Jan 8 at 21:46




$begingroup$
No, the series converges since the measure theory convention, especially in cases like this, is $0 cdot infty = 0$. (But on the other hand, if you count that as divergent, then $f(x) = frac{1}{x^2} chi_{[1, infty)}(x)$ would form a counterexample to the forward direction with the same $m(E_k)$.)
$endgroup$
– Daniel Schepler
Jan 8 at 21:46












$begingroup$
@DanielSchepler Oh, I see. So the statement is actually false. They are not equivalent at all. But if we take the convention $0cdotinfty=0$, then is the above proof for the forward direction correct?
$endgroup$
– Alex
Jan 8 at 21:57




$begingroup$
@DanielSchepler Oh, I see. So the statement is actually false. They are not equivalent at all. But if we take the convention $0cdotinfty=0$, then is the above proof for the forward direction correct?
$endgroup$
– Alex
Jan 8 at 21:57




1




1




$begingroup$
I think a correct statement could go something like: if $F_k = { x mid frac{1}{k+1} < f(x) le frac{1}{k} }$, then $f$ is Lebesgue integrable if and only if $sum_{k=1}^infty k m(E_k)$ and $sum_{k=1}^infty frac{1}{k} m(F_k)$ are both convergent. (In other words, there are two main concerns for integrability: that not too much of a function is "large", and that how much of a function is "small" increases sufficiently slowly.)
$endgroup$
– Daniel Schepler
Jan 8 at 22:05




$begingroup$
I think a correct statement could go something like: if $F_k = { x mid frac{1}{k+1} < f(x) le frac{1}{k} }$, then $f$ is Lebesgue integrable if and only if $sum_{k=1}^infty k m(E_k)$ and $sum_{k=1}^infty frac{1}{k} m(F_k)$ are both convergent. (In other words, there are two main concerns for integrability: that not too much of a function is "large", and that how much of a function is "small" increases sufficiently slowly.)
$endgroup$
– Daniel Schepler
Jan 8 at 22:05










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