Consider a sequence of random variables $(X_n)_{n \in \mathbb{N}}$ on a certain probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Usually, almost sure convergence doesn't imply $L^1$ convergence and vice versa.

However, if we can find a random variable $X$ with $X \in L^1$ such that $\vert X_n \vert \leq \vert X \vert$ almost surely, then

$X_n \to 0$ a.s. $\Leftrightarrow$ $X_n \overset{L^1}{\to} 0 $

by the dominated convergence theorem. Is that true?


1 Answer 1


The convergence of the sequence and domination by $X$ can be relaxed to hold only almost surely provided the measure space is complete. If this is the case, or if you deal as you do with finite measure spaces, then necessity holds since uniform integrability of the sequence $X_n$ is enough.

Sufficiency is not true: just look at here. Clearly all the $X_n$ are bounded by an $L^1$ function, but they do not converge almost everywhere.

  • 1
    $\begingroup$ Maybe a problem on my end but the second link is not working. $\endgroup$
    – Michh
    May 3, 2022 at 21:12
  • $\begingroup$ Thanks! Now it should work $\endgroup$
    – Son Gohan
    May 3, 2022 at 22:00

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