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Consider a probability space $( \Omega, \Sigma, \mu) $ (we could also consider a general measure space). Suppose $f: \Omega -> \mathbb{R}$ is integrable. Does this mean that $ \int |f| \chi(|f| >K) d\mu $ converges to 0 as K goes to infinity? N.B. $\chi$ is the characteristic/indicator function. I showed that if $f$ belongs to $L^2$ as well then we can use the Cauchy-Schwarz inequality and the Chebyshev inequality to show that this is indeed so. For the general case, I think that it is false, but I can't think of a counterxample. I tried $ \frac{1}{\sqrt{x}}$ on $[0,1]$ with the Lebesgue measure, but it didn't work. I can't think of another function that belongs to $L^1$ but not $L^2$! Could anyone please help with this by providing a counterexample or proof? Many thanks.

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Consider approximation by simple fonctions, and then employ the diagonal argument. You could also use the monotone convergence theorem. – William Mar 18 '13 at 0:31
@William "Consider approximation by simple fonctions, and then employ the diagonal argument"... Great Scott! What for? – Did Mar 27 '13 at 15:56
Huh? I am really confused by the above comment. I don't really get what is meant by it. – William Mar 28 '13 at 3:16
@William That I do not understand the hint (the order, really) you provided in the first sentence of your first comment. (Unrelated: if your comment is addressed to some user nameofuser and you want them to be notified of it, write @nameofuser at the beginning of said comment. If you don't, don't.) – Did Mar 29 '13 at 7:13
@Did: My comment is the short (and probably vague) version of the answer provided below by saz. – William Mar 29 '13 at 23:22

Let $f \in L^1$, then we have $|f| \cdot \chi(|f|>k) \leq |f| \in L^1$ and $$|f| \cdot \chi(|f|>k) \downarrow |f| \cdot \chi(|f|=\infty)$$ (i.e. it's decreasing in $k$) since $\chi(|f|>n) \leq \chi(|f|>m)$ for all $m \leq n$. Thus, we obtain by applying dominated convergence theorem $$\lim_{k \to \infty} \int |f| \cdot \chi(|f|>k) \, d\mu = \int |f| \cdot \chi(|f|=\infty) \, d\mu = 0$$ since $\mu(|f|=\infty)=0$.

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Actually one needs $f\in L^1$ to guarantee that the functions $f_k=|f|\chi(|f|\gt k)$ are dominated (by the integrable function $|f|$), not simply that $|f|$ is finite almost everywhere. At the moment this answer is misleading (note that MCT works for nondecreasing sequences, here the sequence $(f_k)$ is nonincreasing...), please consider rewriting it. – Did Mar 18 '13 at 10:56
@Did You are right. I rewrote it. – saz Mar 18 '13 at 13:43

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