Let $(X_i(x))_{i\in\mathbb{N}}$ be a sequence of real random variables with $\operatorname{E}[X_i(x)]=\mu(x)\leq\infty$, for all $i\in\mathbb{N}$ and $x\in\mathbb{R}$. Let's assume that we have $$\sup_{x\in\mathbb{R}}\|X_i(x)\|_{L^1(\Omega)}\xrightarrow{n\rightarrow\infty}0.$$ Is it possible to follow that $$\sup_{x\in\mathbb{R}}|X_i(x)|\xrightarrow{p}0,~~\text{as}~~n\rightarrow\infty?$$

  • $\begingroup$ By $\|X_i(x)\|_{L^1(\Omega)}$ do you mean something like $\int |X_i(x)(\omega)| d \mu(\omega)$? And by $|X_i(x)|$ do you mean $|X_i(x)(\omega)|$? $\endgroup$ – copper.hat Feb 4 '14 at 7:36
  • $\begingroup$ @copper.hat : Yes, for $\sup_{x\in\mathbb{R}}\|X_i(x)\|_{L^1(\Omega)}$ that's what I mean. By $\sup_{x\in\mathbb{R}}|X_i(x)|\xrightarrow{p}0$ I mean: For all $\epsilon>0$ it holds $$\sup_{x\in\mathbb{R}}P(\{\omega\in\Omega:|X_i(x)(\omega)|>\epsilon\})~~\text{tends to}~~ 0,$$ as $n\rightarrow\infty$. $\endgroup$ – stroem Feb 4 '14 at 8:28
  • $\begingroup$ I fixed my answer. $\endgroup$ – copper.hat Feb 4 '14 at 8:51

Let $\epsilon>0$, and $A_i(x) = \{ \omega | |X_i(x)(\omega)| > \epsilon\}$.

Since $|X_i(x)| \ge \epsilon 1_{A_i(x)} $ you have $\int |X_i(x)| \ge \epsilon P(A_i(x))$, and so $P(A_i(x)) \le {1 \over \epsilon} \int |X_i(x)| \le {1 \over \epsilon} \sup_x \int |X_i(x)|$.

Consequently, you have $\sup_x P(A_i(x)) \le {1 \over \epsilon} \sup_x \int |X_i(x)|$, from which the result follows.

  • $\begingroup$ Thanks a lot for the fast and understandable answer! $\endgroup$ – stroem Feb 4 '14 at 8:55
  • $\begingroup$ Delighted to help! $\endgroup$ – copper.hat Feb 4 '14 at 8:59

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