Pratt's Lemma: Let $X_n \leq Y_n \leq Z_n, n=1,2 \cdots$. Suppose with probability $1$, $X_n\rightarrow X, Y_n\rightarrow Y \ \rm{and} \ Z_n\rightarrow Z $. Show that if $E[Z_n]\rightarrow E[Z]$ and $E[X_n]\rightarrow E[X]$, then $E[Y_n]\rightarrow E[Y]$.

Proof: $Z_n-Y_n\geq 0$ and $Y_n-X_n\geq 0$,

By Fatou's Lemma, $E(Z-Y)\leq \liminf E(Z_n-Y_n)$, hence $\limsup E(Y_n)\leq E(Y)$.

Similary,$E(Y)\leq \liminf E(Y_n)$. Thus, $\lim E(Y_n)=E(Y)$.

What I don't understand is that $X$ and $Z$ are not given to be integrable, so how can we cancel $E(Z)$ and $E(X)$?

  • $\begingroup$ The claim is probably false without assuming that $X,Z $ are integrable. I would suggest searching for a counter example. $\endgroup$
    – PhoemueX
    Nov 16, 2016 at 20:12
  • $\begingroup$ This is a question in the book "Knowing the Odds" by John Walsh. I also came across this question in another book so I assumed that the question is right. $\endgroup$ Nov 16, 2016 at 22:29
  • 1
    $\begingroup$ @PhoemueX Yes, if we do not assume $X,Z$ being integrable we cannot eliminate the non-tight situation. See my proof below. $\endgroup$
    – Henry.L
    Nov 30, 2016 at 3:47

1 Answer 1


The proof should follow as this. And we MUST assume that $X$ and $Z$ are both $L^1$ in order to prove that. A counter example is

$Z_n=3n\circ 1_{\{{0<x\leq\frac{1}{n}}\}}$,$Y_n=2n\circ 1_{\{{0<x\leq\frac{1}{n}}\}}$,$X_n=n\circ 1_{\{{0<x\leq\frac{1}{n}}\}}$

Then $Z_n\rightarrow0$ and $Y_n\rightarrow0$ and $X_n\rightarrow0$ poitnwisely. However $EY_n=2\neq 0$. In fact many counterexamples can be found as long as the sequence of measures corresponding to $Z_n$ are not tight.

Proof:$Z_n-Y_n\geq0$ is a sequence of nonnegative random variables by assumption $Y_n\leq Z_n$. And the sequence is pointwisely convergent to $Z-Y$ by assumption. Fatou's lemma said that $$E(\lim \inf Z_n-Y_n)\leq \liminf E(Z_n-Y_n)=\liminf E(Z_n-Y_n)=\liminf E(Z_n)-\liminf E(Y_n)=EZ-\liminf E(Y_n)$$ (1)

But at the same time $$E(\lim inf Z_n-Y_n)=E(\lim Z_n-Y_n)=E(Z-Y)=EZ-EY$$ (2)

By (1) and (2) we have $$EZ-EY\leq EZ-\lim E(Y_n)$$(3) and since $EZ<\infty$ we have $EY\geq \liminf E(Y_n)$. Apply Fatou's Lemma again $$EY=E(\liminf Y_n)\leq \liminf E(Y_n)$$(4)

(3) and (4) result in $EY=\liminf E(Y_n)$

Symmetric argument involving $X_n$ will lead to $EY=\limsup E(Y_n)$ so $\lim E(Y_n)=EY$ is proved. $\square$

This lemma can actually be used to prove Dominated convergence theorem.

Another (possibly) easier proof is that by noticing $L^1$-convergence is equivalent to $L^1$-Cauchy sequence. Pratt's lemma reveals the essence of "dominance", the dominance family passes the bound on $L^1$ norm to the $X_n$ via Fatou's lemma.


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