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Several results (e.g in probability theory or using prob. theory) are stated in an almost surely phrasing (meaning the set of outcomes where this is not so has measure zero)

How can one pass from such statements to the surely mode of such a result?

For example it is kown that some results that hold almost surely hold surely as well (or is strogly conjecture to be the case).

Can one go from the one to the other and if so what additional conditions would be needed?

UPDATE:

As mentioned in the comments a suggestion is that if an almost sure result holds under any possible choice of measure could this imply => sure, since it would not make any practical difference for any measure choice. Sth along these lines..

Is this correct and if so, can this be formalized?

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    $\begingroup$ No, one usually can not. A real number in $[0,1]$ is almost surely irrational, but there are rational numbers in $[0,1]$. $\endgroup$
    – Michael Greinecker
    Jun 5, 2014 at 23:47
  • $\begingroup$ True, however as stated in the question there are many cases where the almost sure result is also sure. What additional conditions could be needed in order to handle this? $\endgroup$
    – Nikos M.
    Jun 5, 2014 at 23:49
  • $\begingroup$ Of course, every statement that is surely true is almost surely true. But there is no general converse. $\endgroup$
    – Michael Greinecker
    Jun 5, 2014 at 23:53
  • $\begingroup$ Yes this is my knowledge so far too. However is this sth formal or just an intuition. For example Hardy's result on the Riemman Hypothesis could be used in such a way to prove the whole RH, sth like this $\endgroup$
    – Nikos M.
    Jun 5, 2014 at 23:55
  • $\begingroup$ The key point (which, I am sorry to say, seems to be missed by every answer posted so far) is that, in most probabilistic settings, an "almost sure" behaviour is the best one can expect. Looking for results concerning an i.i.d. sequence of Bernoulli random variables for example, note that one can realize them on the space $\Omega=[0,1]$ as the bits of the dyadic expansion of $U:\Omega\to[0,1]$, $\omega\mapsto U(\omega)=\omega$. But also as the bits of the expansion of $V(\omega)=\omega$ if $\omega$ is irrational and $V(\omega)=0$ if $\omega$ is rational. These two models of an "i.i.d. .../... $\endgroup$
    – Did
    Jun 15, 2014 at 7:26

2 Answers 2

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"almost sure, ... meaning the set of outcomes where this is not so has measure zero."

To then get to "surely," one must show that the set of outcomes "where this is not" so is the empty set. What else could the answer be?

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  • $\begingroup$ +1, exactly the point of the question! the empty set which has a measure of zero of course. Can this be formalised somehow? $\endgroup$
    – Nikos M.
    Jun 6, 2014 at 0:00
  • $\begingroup$ Suppose that the outcome occurs, and derive a contradiction. No contradiction results from supposing that one randomly picked a number from the interval $[0,1]$ and it came up rational. But a contradiction could be derived from the supposition that the number was greater than 3. $\endgroup$
    – StumpyLeg
    Jun 6, 2014 at 0:04
  • $\begingroup$ hmm yeah i was thinking of maybe in terms of invariant measures or sth like that. So if this holds for an arbitrary measure chosen maybe could imply => surely. Sth along these lines.. $\endgroup$
    – Nikos M.
    Jun 6, 2014 at 0:07
  • $\begingroup$ Let me elaborate on this. If sth holds almost surely for any measure one could use for the set under question, wouldnt this imply => surely, since no other choice of measure could practically make any difference? $\endgroup$
    – Nikos M.
    Jun 6, 2014 at 0:09
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    $\begingroup$ You may want to think of an "almost sure" property $\phi$ that holds for a given set $X$ as one where $\{x \in X \mid \neg \phi(x) \}$ is an element of a given ideal $I \subset X$ rather than in terms of measures. This gives a very handy concept and may convince you that there is no hope to strenghten a result concerning such a property to hold "surely" in general. $\endgroup$
    – Stefan Mesken
    Jun 6, 2014 at 0:17
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There's a basic difference between an "almost sure" event and a "sure" event.

An "almost sure" event is something that will happen with probability 1, even though other events are possible outcomes. For example, if you toss a coin an infinite number of times, it is almost sure that a tails will come up at some point. The event HHHHHHH.... is possible, but as the probability of that goes to 0 the more coin flips you have, then the probability that a tails will come up at some point is 1.

A "sure" event is something that's guaranteed to happen, no matter what. There are no possible outcomes outside of the event. You generally can't "change" an almost sure event to a sure event, because that would require restricting the event in some new way that takes the outcomes that are technically possible but have probability zero and makes them completely impossible.

You don't truly have things that are both "almost sure" and "sure". It's often easier to prove that something is "almost sure" than it is to prove it's "sure", but once you prove it's "sure" it's no longer "almost sure".

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  • $\begingroup$ What you state is correct, except one point. The cases where the almost sure result stems from lack of data or knowledge to prove the sure event (which in many cases is known to hold). So almost sure in this example means "almost there" to use a play on words ;) $\endgroup$
    – Nikos M.
    Jun 5, 2014 at 23:58
  • $\begingroup$ If the probability of an event is not just infinitesimally close to zero but strictly zero, then it also "very close to zero." In that sense "surely" implies "almost surely" as Michael G suggested. $\endgroup$
    – StumpyLeg
    Jun 6, 2014 at 0:00

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