Solution: Probability Zero Is Not Impossible

Exercise: Probability Zero Is Not Impossible

Part 1

The event $\{U = 0.5\} = \{0.5\}$ is a singleton in $[0, 1]$ . Under Lebesgue measure:

\mathbb{P}(U = 0.5) = \mathbb{P}(\{0.5\}) = \mathbb{P}([0.5, 0.5]) = 0.5 - 0.5 = 0

The event has probability zero, but it is not impossible. If you sample from

U \sim \text{Uniform}[0,1]

, the outcome

U = 0.5

is a genuine element of

\Omega = [0, 1]

— it is in the sample space and can in principle occur. It simply has measure zero.

Part 2

For any $x \in [0, 1]$ : $\mathbb{P}(U = x) = 0$ by the same argument. Every individual outcome in a continuous probability space has probability zero. The probability measure assigns positive mass only to intervals (or more general sets with positive Lebesgue measure), never to individual points.

This is the defining feature that distinguishes continuous from discrete probability: in a discrete space, individual outcomes have positive probability; in a continuous space, all the probability is "spread across" uncountably many outcomes.

Part 3

The trader's reasoning is flawed.

\mathbb{P}(S_T = K) = 0

is a fact about a single point, but it says nothing about the density of the distribution near

K

. What matters for option pricing near expiry is the probability density

f_{S_T}(K)

, or equivalently the probability that

S_T

lands in a small interval around

K

\mathbb{P}(K - \epsilon < S_T \leq K) = \int_{K-\epsilon}^{K} f_{S_T}(s)\,ds > 0

for any

\epsilon > 0

. A call struck at

K

that expires near the money has significant delta and gamma precisely because there is a non-negligible probability of landing in any interval straddling

K

, even though the probability of landing exactly at

K

is zero.

The practically relevant quantity is the probability of the option expiring in-the-money:

\mathbb{P}(S_T > K) = \Phi(d_2)

in the Black-Scholes formula — a genuinely positive number.

Part 4

Neither implication holds in general.

$\mathbb{P}(A) = 0$ does not imply $A = \emptyset$ . As shown above, $\{0.5\}$ has probability zero but is non-empty. Any countable subset of $[0, 1]$ also has Lebesgue measure zero. Events with probability zero are called null events or $\mathbb{P}$ -null sets.
$\mathbb{P}(B) = 1$ does not imply $B = \Omega$ . Take $B = (0, 1]$ . Then $\mathbb{P}(B) = 1$ under Lebesgue measure, but $B \neq [0, 1] = \Omega$ since $0 \notin B$ . Events with probability one are said to hold almost surely (a.s.), not necessarily everywhere.

The correct implications run the other way: $A = \emptyset \Rightarrow \mathbb{P}(A) = 0$ , and $B = \Omega \Rightarrow \mathbb{P}(B) = 1$ . But the converses fail in any continuous model.

Takeaways

In continuous probability spaces, every individual outcome has probability zero. "Probability zero" means measure-theoretically negligible, not logically impossible.
The practically relevant quantities are probabilities of intervals (or measurable sets with positive measure), captured by CDFs and densities — not point probabilities.
Almost surely (a.s.) is the correct qualifier for statements that hold on a set of probability one but not necessarily everywhere. Stochastic calculus is full of a.s. statements: Itô integrals are defined a.s., solutions to SDEs exist a.s., and so on.
The distinction matters in finance: a digital option with payoff $\mathbf{1}_{S_T = K}$ is worth exactly zero in any diffusion model, while a digital with payoff $\mathbf{1}_{S_T > K}$ is worth $e^{-rT}\mathbb{Q}(S_T > K) > 0$ .