Uniform Probability on Intervals

A Uniform Point on a Finite Interval

To model a point chosen uniformly from a finite interval, the probability assigned to a measurable set should depend on how much of the interval it occupies, not on how many individual points it contains. Lebesgue measure provides the notion of length, and dividing by the total length makes that length into a probability. This gives exact probabilities for subintervals, unions of intervals, and many events described by distance conditions.

Fix real numbers $a<b$. Use the measurable space $([a,b],\mathcal{L}_{[a,b]})$, where $\mathcal{L}_{[a,b]}$ consists of the sets $E\cap[a,b]$ for $E$ Lebesgue measurable in $\mathbb{R}$. Write $m$ for Lebesgue measure on $\mathbb{R}$. This space has finite measure $m([a,b])=b-a$, and $\mathcal{L}_{[a,b]}$ is complete.

Definition: The uniform probability measure on $[a,b]$ is the set function $P$ on $\mathcal{L}_{[a,b]}$ defined by $$ P(E)=\frac{m(E)}{b-a},\qquad E\in\mathcal{L}_{[a,b]}. $$ Thus $([a,b],\mathcal{L}_{[a,b]},P)$ is a probability space: its total probability is $1$, and measurable events are assigned their Lebesgue length divided by the length of the whole interval. In this model, “a uniformly chosen point” means that an event is a Lebesgue measurable subset of $[a,b]$, with probability given by this formula.

Why Normalizing Length Gives a Probability

Proposition: For $a<b$, the set function $P(E)=m(E)/(b-a)$ on $\mathcal{L}_{[a,b]}$ is a complete probability measure.

Proof. Since $m$ is a measure and $b-a>0$, $P$ is nonnegative and $P(\varnothing)=m(\varnothing)/(b-a)=0$. Also $P([a,b])=m([a,b])/(b-a)=(b-a)/(b-a)=1$. If $E_1,E_2,\ldots$ are pairwise disjoint members of $\mathcal{L}_{[a,b]}$, countable additivity of Lebesgue measure gives $$ P\left(\bigcup_{j=1}^{\infty}E_j\right) =\frac{m\left(\bigcup_{j=1}^{\infty}E_j\right)}{b-a} =\frac{\sum_{j=1}^{\infty}m(E_j)}{b-a} =\sum_{j=1}^{\infty}P(E_j). $$ Finally, if $N\in\mathcal{L}_{[a,b]}$ has $P(N)=0$, then $m(N)=0$. Every subset of $N$ is Lebesgue measurable by completeness of Lebesgue measure and has measure zero; it therefore also has $P$-measure zero. This proves completeness. $\square$

The interval formula follows directly from the definition. It applies to Lebesgue measurable events, not just intervals. For an interval event, its endpoints do not affect the answer: a single point has Lebesgue measure zero. Indeed, for any $c\in\mathbb{R}$ and every $\varepsilon>0$, $\{c\}\subset(c-\varepsilon,c+\varepsilon)$, so $0\le m(\{c\})\le 2\varepsilon$. Since this holds for every positive $\varepsilon$, $m(\{c\})=0$.

Theorem: Let $([a,b],\mathcal{L}_{[a,b]},P)$ be the uniform probability space with $a<b$. If $a\le u\le v\le b$, then any interval with endpoints $u$ and $v$, formed using open or closed endpoints and contained in $[a,b]$, has probability $(v-u)/(b-a)$, except that an interval with $u=v$ may be empty or a singleton, both of probability zero.

Proof. If $u<v$, the interval $(u,v)$ has Lebesgue measure $v-u$. Adding either endpoint, or both, changes it by a subset of $\{u,v\}$. Each endpoint has measure zero, so every such endpoint choice has measure $v-u$, and its probability is $m(J)/(b-a)=(v-u)/(b-a)$. This also covers intervals that use $a$ or $b$ as an endpoint, since the same singleton argument applies at the boundary. If $u=v$, an interval with these endpoints is either empty or $\{u\}$; each has measure zero and hence probability zero. $\square$

Worked Example: A Half-Open Subinterval

Let the sample space be $[2,10]$, with its Lebesgue measurable sets and uniform probability $P(E)=m(E)/(10-2)$. Find the probability that the chosen point lies in $E=(3,7]$.

The whole interval has length $10-2=8$. The event interval has length $7-3=4$, so $$ P((3,7])=\frac{m((3,7])}{m([2,10])} =\frac{7-3}{10-2} =\frac{4}{8} =\frac12. $$ The open endpoint at $3$ and the closed endpoint at $7$ do not change the length. Check: the complement in $[2,10]$ has length $(3-2)+(10-7)=1+3=4$, so its probability is $4/8=1/2$. The event and its complement therefore have probabilities $1/2+1/2=1$.

Unions of Events and Overlap

For disjoint measurable events, additivity says that the probability of their union is the sum of their probabilities. If the events overlap, adding their lengths counts the shared part twice. One must either split the union into disjoint pieces or subtract the overlap. For two measurable events $A$ and $B$, finite additivity gives $P(A\cup B)=P(A)+P(B)-P(A\cap B)$: the disjoint pieces $A\setminus B$, $B\setminus A$, and $A\cap B$ partition $A\cup B$.

Worked Example: A Union with an Overlap

In the uniform model on $[0,12]$, take $A=[1,7]$ and $B=[5,10)$. Compute $P(A\cup B)$.

The total interval has length $12-0=12$. The lengths are $m(A)=7-1=6$ and $m(B)=10-5=5$. Their overlap is $[5,7]$, of length $7-5=2$. Therefore, $$ m(A\cup B)=6+5-2=9,\qquad P(A\cup B)=\frac{9}{12}=\frac34. $$ Equivalently, the union is $[1,10)$, whose length is $10-1=9$. Check: the complement in $[0,12]$ consists, up to endpoints, of lengths $1-0=1$ and $12-10=2$. Its total length is $1+2=3$, so $P((A\cup B)^c)=3/12=1/4$, and $3/4+1/4=1$.

Distance Conditions as Events

A geometric description becomes a probability calculation once it is translated into a subset of the interval. For example, a condition $|x-c|\le r$, where $r\ge0$, selects the points in $[c-r,c+r]$, restricted to the sample interval. The favorable length is the length of that intersection with the sample interval. If several distance conditions are joined by “or,” take the union of the corresponding subsets and account for any overlap.

Worked Example: Within a Given Distance of Either Mark

Choose a point uniformly from $[0,12]$. Find the probability that it is within distance $2$ of $3$ or within distance $2$ of $8$. Here “within distance” uses the condition $\le 2$.

The first condition is $|x-3|\le2$, equivalent to $1\le x\le5$, so its event is $[1,5]$, of length $5-1=4$. The second is $|x-8|\le2$, equivalent to $6\le x\le10$, so its event is $[6,10]$, also of length $10-6=4$. These intervals are disjoint because $5<6$. Their union therefore has length $4+4=8$, and $$ P\bigl([1,5]\cup[6,10]\bigr)=\frac{8}{12}=\frac23. $$ Check: the complement has three pieces, with lengths $1-0=1$, $6-5=1$, and $12-10=2$. The total complementary length is $1+1+2=4$, giving complement probability $4/12=1/3$; thus the event probability is $1-1/3=2/3$.

Common Pitfalls

Counting endpoints instead of measuring length. A point in a continuous interval is not equally likely to be any one of a finite list of points: the model assigns probability according to Lebesgue length. For instance, on $[0,1]$, the event $\{1/2\}$ has probability $m(\{1/2\})/m([0,1])=0/1=0$, not $1/2$ or $1$. An interval’s endpoint conventions also do not change its probability, because each endpoint has measure zero.

Adding overlapping lengths without correction. For the events $A$ and $B$ in the preceding union example, the incorrect calculation $P(A\cup B)=(6+5)/12=11/12$ counts the overlap of length $2$ twice. The correction is $$ P(A\cup B)=\frac{6+5-2}{12}=\frac{9}{12}=\frac34. $$ A reliable alternative is to draw or describe the disjoint pieces of the union and add their lengths once each.

The normalization also makes comparisons across intervals straightforward: a measurable event occupying a fraction $q$ of the total interval has probability $q$. The fraction must use the length of the favorable subset, with overlaps removed and portions outside the sample interval excluded. No assumption that the event itself is an interval is needed; only Lebesgue measurability is required by the definition.

Check Your Understanding

For each question, specify the interval model and use length divided by total length. Pay attention to endpoint choices and overlap.

In the uniform model on $[1,9]$, compute the probability of the event $(2,6)$.
In the uniform model on $[0,10]$, find the probability of $[1,5]\cup[4,8]$. Identify the overlap before calculating.
On $[0,14]$, find the probability that a uniformly chosen point is within distance $1$ of $4$ or within distance $2$ of $10$, using strict inequalities for both distance conditions.
Explain why the probabilities of $[u,v]$ and $(u,v)$ agree when $a\le u<v\le b$.
Give an example of a measurable event in $[0,1]$ that has probability zero, and compute its probability from the definition.