Question about integration (related to uniform integrability)

Question

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.

Answer 1

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 $\mu(|f|=\infty)=0$.