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Assume $X$ is a measure space and $f : X \to [0,\infty]$ is an absolutely integrable function (that is $\int_X f \; d \mu < \infty$). This question is about the asymptotic behaviour of the function $$E(\delta) = \int _{f \leq \delta} f \; d \mu.$$

Since $f$ is absolutely integrable, it follows from the downward monotone convergence theorem that $E(\delta) \to 0$ as $\delta \to 0$. What I want to understand is how fast $E(\delta)$ converges to 0.

What I know is that if $\mu(X) < \infty$ then we have $E(\delta) \leq \delta \mu(X)$, so $E(\delta)$ converges to $0$ at a linear rate or faster.

Question: What can be said when $\mu(X) = \infty$?

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up vote 3 down vote accepted

You can't say anything in general. For a simple example showing that you can't get any polynomial rate, consider the measure space $X = [1, \infty)$ with the Lebesgue measure and the function $f(x) = x^{-(1 + \epsilon)}$. Assuming I integrated it correctly, you get $E(\delta) = \frac{1}{\epsilon} \delta^{\epsilon/(1 + \epsilon)}$ which gets arbitrarily slow as $\epsilon \to 0$.

You can also do a more complicated construction that gives a bit more: for any function $e(\delta)$ with $e(\delta) \to 0$, you can find an $f$ such that $E(\delta_k) \ge e(\delta_k)$ for some sequence $\delta_k \to 0$. In particular, you can get arbitrarily slow rates. For this construction, choose a decreasing sequence $\delta_k \to 0$ with $e(\delta_k) \le 2^{-k}$. Then define $f$ such that $\mu(f = \delta_k) = 2^{-k}/\delta_k$. (If $X$ is $[0, \infty)$ with Lebesgue measure, you can do this by setting $f(x) = \delta_k$ if $\sum_{j \le k-1} 2^{-j}/\delta_j \le x < \sum_{j \le k} 2^{-j}/\delta_j$.) Then $f$ is integrable because $$ \int f = \sum_{k} \delta_k \mu(f = \delta_k) = \sum_k 2^{-k} < \infty. $$ But you have $$ E(\delta_k) = \int_{f \le \delta_k} f = \sum_{j \ge k} \delta_j \mu(f = \delta_j) \ge \delta_k \mu(f = \delta_k) = 2^{-k} = e(\delta_k). $$

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