This question has been addressed previously in here but my question is different. It is from Hogg and McKean's "Introduction to Mathematical Statistics".

Theorem $5.2.7.$ Let $\{X_n\}$ be a sequence of random variables bounded in probability and let $\{Y_n\}$ be a sequence of random variables that converges to $0$ in probability. Then $$X_n Y_n \xrightarrow{P} 0 .$$

Proof: Let $\epsilon > 0$ be given. Choose $B_{\epsilon} > 0$ and an integer $N_{\epsilon}$ such that $n \geq N_{\epsilon} \implies P[|X_n| \leq B_{\epsilon}] \geq 1−\epsilon.$

Then $$ \begin{align} \overline{\lim_{n \to \infty}} P[|X_nY_n|\geq \epsilon] &\leq \overline{\lim_{n \to \infty}} P[|X_nY_n|\geq \epsilon,|X_n|≤B_{\epsilon}] \\ &+ \overline{\lim_{n \to \infty}} P[|X_nY_n| \geq \epsilon,|X_n| > B_{\epsilon}] \\ &\leq \overline{\lim_{n \to \infty}} P [|Y_n| ≥ \epsilon/B_{\epsilon}] + \epsilon = \epsilon. \end{align}$$ from which the desired result follows.

How exactly does the desired result simply follow?

In the proof above, $\overline{\lim}$ represents the limit supremum (limsup) of that sequence. My issue with this proof is that they show that $$\overline{\lim_{n \to \infty}} P[|X_nY_n|\geq \epsilon] \leq \epsilon$$ but what we want to show is that there exists $N$ such that $$\sup_{n \geq N} \{P[|X_nY_n|\geq \epsilon]\} \leq \delta$$ for any given $\delta$ unrelated to $\epsilon$, which can be made arbitrarily small. The variable $\epsilon$ appears as part of the condition for boundedness of $X_n$ in probability. I don't understand how the statement that the limsup is less than $\epsilon$ would imply that the limsup actually goes to zero. Please help me understand the proof clearly.


2 Answers 2


The following general result would come handy in deducing what the authors did above.

Result $1.$ Let $(\Omega,\boldsymbol{\mathfrak A},\mu)$ be a measure space. Let $\langle f_n\rangle_{n\in\mathbb N}$ be a sequence of extended real-valued $\boldsymbol{\mathfrak A}$-measurable functions on $D\in\boldsymbol{\mathfrak A};$ let $f$ also be an extended real-valued $\boldsymbol{\mathfrak A}$-measurable function on $D.$ Then $$f_n\overset{\mu}{\longrightarrow} f\iff \forall ~\delta > 0~\exists ~N_\delta\in\mathbb N:\mu\{D:\vert f_n-f\vert\geq\delta\}<\delta~~\forall~ n\geq N_\delta.\tag 1\label 1$$

To see $\eqref 1\implies ~\mu-\textrm{convergence}, $ note that you can take $\delta:=\eta\wedge\varepsilon$ where both $\eta, ~\varepsilon> 0$ and are arbitrarily chosen. By consideration of such $\delta, $ it implies $\{D:\vert f_n-f\vert\geq\varepsilon\} \subset\{D:\vert f_n-f\vert\geq\delta\}$ following which we get $$\mu\{D:\vert f_n-f\vert\geq\varepsilon\} \leq \mu\{D:\vert f_n-f\vert\geq\delta\}<\delta\leq \eta~~\forall~ n\geq N_\delta\implies \mu\{D:\vert f_n-f\vert\geq\varepsilon\}<\eta~~\forall~ n\geq N_\delta,$$ which is nothing but $\lim_{n\to\infty}\mu\{D:\vert f_n-f\vert\geq\varepsilon\}=0, $ the definition of convergence in measure.


You can now formally show how it is related to the author's statement.


$\rm [I]$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, $2014, $ sec. $1\S6\rm[III], $ pp. $111-112.$

  • $\begingroup$ Thanks for the answer. It is proposition 6.17 in Yeh's book. It took me a while to figure out what $\eta \wedge \varepsilon$ meant. Then I found this link on math SE math.stackexchange.com/q/3454725/145325. So essentially, we apply the logic to minimum of $\eta,\varepsilon$ after which I can say, like the author did, that "the desired result follows." :) :) $\endgroup$ Commented Oct 31, 2023 at 8:30
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    $\begingroup$ Yeh's book is very much underrated and yet one of the most comprehensive books, and I would recommend it to anyone. $\endgroup$ Commented Oct 31, 2023 at 8:34

If the $\limsup$ is less than $\epsilon$ for every positive $\epsilon$, then it's zero. It can't be, for example, $1/3, $ because you could take $\epsilon=1/4$. It can't be $0.00042,$ because you could take $\epsilon=0.00001$. It can't be $2^{-10^{100}}$, because you could take $\epsilon=2^{-10^{100}-1}$. And so on.

You don't need a $\delta$ separate from $\epsilon$, it doesn't add any more generality.


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