I am trying to understand the implication of the Borel-Cantelli Lemma to the random oracle model.

I think understanding a special case, say, a random oracle is one-way with probability 1, would be helpful. The statement (see, e.g., page 19 of Arkady Yerukhimovich's thesis) as far as I understand in words goes like "if the adversary $A$ given access to an $n$-bit random oracle $O_n$ succeeds in the game (given $y$, output a preimage $x\in O^{-1}(y)$) with probability at most $1/n^2$, then $A$ must fail the game for sufficiently large $n$ with probability 1 over choices of $O$."

I don't understand what it means by "with probability 1 over choices of $O$," when $O$ does not refer to a specific size $n$ (which I think is the setting where the Borel-Cantelli Lemma applies). The adversary $A$ is uniform, and therefore the adversary $A_n$ for $n$-bit oracle can be constructed by learning the size of $O$.

Let the distribution $D_n$ be the uniform distribution over $n$-bit functions. Clearly $D_n$ is the distribution of an $n$-bit random oracle. But I suspect that the same statement works for other distribution (i.e., whether the oracle distribution is uniform or not does not matter).

Does the statement mean that for $D=D_1\times D_2\times\ldots$ and a sequence of oracles $O_1,O_2,...$ sampled from $D$, $A_1,A_2...$ must fail the game for sufficiently large $n$ with probability $1$ over choices $O_1,O_2,...$? If that's correct, then I'm not sure if $D$ is well-defined. If not, I'd appreciate if you could let me know precisely what the correct statement is.

  • $\begingroup$ Borel cantelli lemma is hardly a standard cryptography tool. You can improve your question by stating it. Also, does it not apply to infinite sequence of RVs? $\endgroup$
    – kodlu
    Commented Dec 9, 2022 at 2:19
  • 1
    $\begingroup$ Thanks. I improved the question. I suspect it applies but I'm not sure. $\endgroup$
    – user50394
    Commented Dec 9, 2022 at 7:13
  • $\begingroup$ "I don't understand what it means by "with probability 1 over choices of $O$," when $O$ does not refer to a specific size $n$". Since the number of possible $O=\{O_1,O_2,\cdots\}$ is uncountable, we have to rely on measure theory. So, that statement should really be "for measure 1 of $O$s". There is a one-to-one correspondence between random (bit) oracles and reals: choose a random real $r$ and then the set the value of $O(x)$ as $r_i$, the $x$-th bit of $r$. So, it boils down to defining measure for reals, which you can find in standard textbooks. Does this clear your doubt? $\endgroup$
    – ckamath
    Commented Dec 10, 2022 at 9:49
  • $\begingroup$ @ckamath I think my confusion is that whether the Borel-Cantelli lemma really means A wins the game with probability/measure 0 over $O_1,O_2,\ldots$. And my interpretation of your answer seems to suggest it is correct. And thanks for clarifying on whether the distribution over $O_1,O_2,\ldots$ can be made well-defined, though I still don't get what it means by choosing a random real. $\endgroup$
    – user50394
    Commented Dec 10, 2022 at 19:03
  • $\begingroup$ "... whether the Borel-Cantelli lemma really means A wins the game with probability/measure 0 over..". It's subtle. It is used to show that only for measure $0$ of $O$s the adversary's advantage is more than expected at infinitely-many $n$s. In other other words, for measure $1$ of $O$s, the adversary's advantage is more than expected at only finitely-many $n$s. This can now be used to argue that for measure $1$ of $O$s, the adversary's advantage is negligible. $\endgroup$
    – ckamath
    Commented Dec 11, 2022 at 20:35

1 Answer 1


$ \newcommand{\AA}{\mathsf{A}} \newcommand{\sO}{\mathcal{O}} \newcommand{\fO}{O} \newcommand{\str}{\{0,1\}^*} \newcommand{\strn}{\{0,1\}^n} \newcommand{\NN}{\mathbb{N}} \newcommand{\adv}{\varepsilon_{\AA,\fO}} \newcommand{\sP}{\mathcal{P}} \newcommand{\SUCCESS}{\text{SUCCESS}_{\AA,\fO,x}} \newcommand{\DEVIATION}{\text{DEVIATION}_{\AA,\fO,n}} $

Let's go through the proof from [Y,IR] in a bit more detail since it is technical and subtle (I struggled a lot with it). The role of Borel-Cantelli lemma will become clear in the process, and is summarised at the end.

Random oracles. We consider (function) oracles $\fO:\str\to\str$ interpreted as an ensemble of oracles $\{O_1,O_2,\cdots\}$, where $\fO_n:\strn\to\strn$. Let $\sO$, denote the set of all such oracles. Since $|\sO|$ is uncountable, before talking about random oracles, we need to define what it means to randomly sample from a sample space that is uncountable. To this end, one defines a probability measure. Since there is a one-to-one correspondence$^*$ between $\sO$ and $[0,1)$, one can resort to Borel sets and Lebesgue measure: see this lecture note for more details.

Random oracles are one-way. Now, our goal is to show that random oracles are one-way in a very strong sense: for measure $1$ of random oracles $\fO$, $\fO$ is a one-way function (OWF), i.e., $$ \Pr_{\fO\leftarrow\sO}[\forall\AA\in\text{PPT}:\adv(\cdot)\text{ is negligible}]=1, $$ where the advantage $\adv(\cdot)$ is defined as $$ \adv(n):=\Pr_{\AA,x\leftarrow\strn}[\underbrace{\AA(\fO(x))\in\fO^{-1}(O(x))}_{\text{Event }\SUCCESS}], $$ and it is negligible if $$ \forall c\in\NN~\exists n_c\in\NN~\forall n>n_c:\adv(n)\geq1/n^c. $$ We proceed as follows.

  1. Let's first analyse the advantage with respect to a fixed adversary and input. It can be shown by lazy sampling$^{**}$ that $$ \forall\AA\in\text{PPT}~\forall n\in\NN~\forall x\in\strn:\Pr_{\fO\leftarrow\sO}[\SUCCESS]\leq n^a/2^n, $$ where, for $a\in\NN$ (which depends on $\AA$), $n^a$ is the upper bound on $\AA$'s runtime.

  2. Next, we bound the probability that $\AA$ deviates from expected behaviour. To this end, let's define a bad event $$ \DEVIATION:\adv(n)> n^{a+2}/2^n. $$ It can be shown by applying Markov's inequality$^{**}$ that $$ \forall\AA\in\text{PPT}\forall n\in\NN:\Pr_{\fO\leftarrow\sO}[\DEVIATION]\leq1/n^2. $$

  3. We are now ready to apply Borel-Cantelli lemma. Since $$ \sum_{n=1}^\infty\Pr_{\fO\leftarrow\sO}[\DEVIATION]\leq \sum_{n=1}^\infty1/n^2<\infty, $$ by Borel-Cantelli lemma, we get that $$ \forall\AA\in\text{PPT}:\Pr_{\fO\leftarrow\sO}[\DEVIATION \text{ occurs for infinitely-many } n \text{s}]=0. $$

  4. This means for each adversary $\AA$, we can fix a measure $0$ of "bad" oracles $\sO^*_\AA\subseteq\sO$. Since the number of PPT Turing machines is countable (but infinite) and union of countable measure $0$ oracles is still measure $0$, we get that the set of all bad oracles $\sO^*=\cup_\AA\sO^*_\AA\subseteq\sO$ is still measure $0$. Therefore, we can switch the order of the quantifiers to get: $$ \Pr_{\fO\leftarrow\sO}[\forall\AA\in\text{PPT}:\DEVIATION \text{ occurs for infinitely-many } n \text{s}]=0. $$

  5. Finally, let's establish one-wayness. The above equation is equivalent to $$ \Pr_{\fO\leftarrow\sO}[\forall\AA\in\text{PPT}:\adv(n)>n^{a+2}/2^n \text{ for finitely-many } n \text{s}]=1. $$ It follows that there exists a $n_\AA\in\NN$ such that $\forall n>n_\AA,\adv(n)\leq n^{a+2}/2^n$. Since $n^{a+2}/2^n$ is a negligible function, and $\adv(n)$ grows slower than $n^{a+2}/2^n$ for all $n>n_\AA$, it follows that $\adv(n)$ is also negligible, which completes the proof.

To sum up, the Borel-Cantelli lemma is used to show that the bad event, $\DEVIATION$, does not occur infinitely-often, which then implies that it does not occur after a sufficiently large $n$, just like in the definition of negligible. This is key to establishing that the advantage is negligible.

$^*$This correspondence is easy to see for bit oracles: given a real number $r=0.r_0r_1\cdots\in[0,1)$, simply set the output $O(x)$ as $r_x$, the $x$-th bit of $r$. This can be extended to the function oracles as shown in [IR].

$^{**}$ See [IR,Y] for details.

[IR]: Impagliazzo and Rudich, Limits on the Provable Consequences of One-Way Permutations. STOC 1989.

[Y]: Yerukhimovich, A Study of Separations in Cryptography: New Results and Models, 2011, PhD Thesis


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