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If $f\in L^1(\mathbb{R})$, I would like to prove or disprove that $$\lim_{\lambda\to\infty}\int_{|f|\ge \lambda}|f|\,dx=0.$$ I believe it is true, though I am not certain, as I have verified it for various integrable functions, such as the set $\{x^{-\alpha}\chi_{[0,1]}: 0<\alpha<1\}\subset L^1(\mathbb{R})$. Moreover, it fails to be true for $x^{-1}\chi_{[0,1]}$, which is indeed not integrable.

The result is of course true if $|f|$ is bounded, and I know that I can approximate arbitrarily well $f$ in $L^1$ by test functions $\psi\in C_c^\infty$ (given $\varepsilon>0$, choose $\psi$ such that $|\psi-f|_{L^1)<\varepsilon$), but I then am left with the following $$\int_{|f|\ge \lambda}|f|\,dx< \varepsilon + \int_{|f|\ge \lambda} |\psi|\,dx$$ and I am not sure how the handle the rightmost term.

Mittens
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Dispersion
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    This has been asked many times before in MSE. From this posting the conclusion follows easily since $||f|>\lambda|\leq\lambda^{-1}\int|f|\xrightarrow{\lambda\infty}0$. More direct proofs of this abound using dominated convergence: $|f|\mathbb{1}{{|f|>a}}|f|\leq|f|$ and $||f|\mathbb{1}{{|f|>a}}\xrightarrow{a\rightarrow\infty}0$ a.s. since $\int|f|<\infty$. – Mittens Jun 20 '23 at 18:33
  • Is the proper etiquette to close this question? A quality answer has been provided which will be lost if I do so. – Dispersion Jun 21 '23 at 22:44
  • Nothing is lost. It is a common and trivial question, however, it would not be nice for you to close it since you already accepted an answer. Should the community decide to close it, it would be though downvoting first and then voting to delete. I will abstain in this case. – Mittens Jun 22 '23 at 05:01

1 Answers1

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It's true (in any measure space $(X,\Sigma;\mu))$. Maybe the easiest way to see this is to consider the sequence $|f|\cdot\chi_{|f|>\lambda_n}$ (where $(\lambda_n)_{n\in\Bbb N}$ is a sequence of positive reals that diverges to $+\infty$) and note this sequence is dominated by $|f|\in L^1$, so the D.C.T (or an easy adapation of the M.C.T) applies to say: $$\lim_{n\to\infty}\int_{|f|>\lambda_n}|f|\,\mathrm{d}\mu=\int_{f^{-1}\{\pm\infty\}}|f|\,\mathrm{d}\mu=0$$

Noting that if $f$ is integrable, $\mu(f^{-1}\{\pm\infty\})=0$ is forced.

An "alternative" method: integration by $f$ is absolutely continuous. From $\mu(f^{-1}\{\pm\infty\})=0$ and $\mu(|f|^{-1}(\lambda_n,\infty))<\infty$ for all $n$ (else, $\int_X|f|\,\mathrm{d}\mu\ge\lambda_n\cdot\mu(|f|^{-1}(\lambda_n,\infty))=\infty$ which contradicts integrability) we can use continuity of measure to infer that $\mu(|f|^{-1}(\lambda_n,\infty))$ can be made arbitrarily small for large $n$. Coupling with absolute continuity of the integral, we can deduce the integral of $|f|$ over the arbitrarily small measure sets $\{|f|>\lambda_n\}$ can also be made arbitrarily small.

This is not really an alternative since iirc absolute continuity of the integral is proven using either the M.C.T or D.C.T, but it's nice.

FShrike
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