What's wrong with my "proof" that the Lebesgue measure of $[0,1]$ is $0$?

Question

Proposition. $\lambda [0,1] = 0$

Proof. Let $\varepsilon>0$ be arbitrary. We will prove that $\lambda[0,1] <\varepsilon$. Let $q : \mathbb{N} \rightarrow \mathbb{R}$ denote an injection with image equal to $\mathbb{Q} \cap [0,1]$. Let $p : \mathbb{N} \rightarrow \mathbb{R}_{>0}$ denote a sequence with $\sum_{i \in \mathbb{N}} p_i < \varepsilon$. Then $$\lambda [0,1] = \lambda \bigcup_{i \in \mathbb{N}} (q_i+[-p_i/2,p_i/2]) \leq \sum_{i \in \mathbb{N}} \lambda (q_i+[-p_i/2,p_i/2]) = \sum_{i\in \mathbb{N}} p_i < \varepsilon$$

So $\lambda[0,1] < \varepsilon$. Since this is true for all $\varepsilon>0$, we deduce that $$\lambda [0,1] = 0.$$

Question. What gives?

Answer 1

Yep. The fact that (in your notation) $$ [0,1] \subset \bigcup_{j=1}^{\infty} (q_j - p_j/2,q_j+p_j/2) $$ is untrue is classic fact one has to get past to begin to understand Lebesgue measure.

By picking $p_j = \epsilon/2^j$, you show exactly that there are open sets containing $\mathbb{Q} \cap [0,1]$ with Lebesgue outer measure at most $\epsilon$. (In particular they are measurable and hence the complement is measurable with measure $> 1 - \epsilon$, so they certainly don't cover $[0,1]$! )

Answer 2

Looking at other answers and the discussions in comments here, it appears that the key issue here is not being handled properly. And previous version of my answer is perhaps not understood well by the community. What follows is an updated version with more explanation.

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First some explanation about the question. The sequence $q_{i}$ is the just arrangement of rational numbers of $[0, 1]$ into a particular sequence (this is possible because rationals form a countable subset). Corresponding to a given $\epsilon > 0$ another sequence $p_{i}$ of positive real numbers is chosen so that $\sum_{i} p_{i} < \epsilon$. OP now argues that since the interval $[0, 1]$ is contained in the union $$\bigcup_{i = 1}^{\infty}[q_{i} - p_{i}/2, q_{i} + p_{i}/2]$$ therefore its measure is not more than the sum of lengths of all these intervals $[q_{i} - p_{i}/2, q_{i} + p_{i}/2]$. And thus the measure of $[0, 1]$ is less than $\epsilon$ and since $\epsilon$ was arbitrary this means that $[0, 1]$ has measure $0$. And OP knows that he/she has committed a mistake somewhere because it is common knowledge that measure of $[0, 1]$ is $1$.

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Most answers try to treat this in a high brow manner by saying that the proof is wrong because a sequence of intervals of total length $a$ cannot cover an interval of length $b$ if $a < b$. This fact appears obvious / intuitive but a proper proof depends on the nature of real numbers and specifically we need some powerful theorem like Heine Borel to handle this. The same statement is trivial to prove if the sequence of intervals considered is finite. It is a common mistake in analysis to extend the finitary arguments to situations where one is dealing with infinite. Most of the times this is possible but it has to be justified by a deeper analysis because in general it is not true. Thus for example finite sets are always countable, but there are infinite sets which are countable and there are also infinite sets which are uncountable.

Another issue is regarding the common knowledge that measure of $[0, 1]$ is $1$. This is the first fundamental theorem in measure theory which again appears intuitive/obvious but requires Heine Borel theorem (or its equivalent) for its proof.

All the answers agree that the problem with OP's proof lies in his assumption $$[0, 1] \subseteq \bigcup_{i = 1}^{\infty}[q_{i} - p_{i}/2, q_{i} + p_{i}/2]\tag{1}$$ and yes this equation is wrong but this can not be proved by using measure theory and obtaining a contradiction like OP has done.

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Let's assume equation $(1)$ and derive a contradiction using Heine Borel Theorem. Since the intervals $[q_{i} - p_{i}/2, q_{i} + p_{i}/2]$ under consideration are closed intervals, it may seem that Heine Borel is not applicable. But this issue is trivial to fix. Clearly we may have a problem only with the boundary points $q_{i} - p_{i}/2, q_{i} + p_{i}/2$ and the problem will occur only if the point is irrational (rationals are already covered by the intervals in equation $(1)$). And this is possible only when $p_{i}$ is irrational. Consider all these points of type $q_{i} - p_{i}/2, q_{i} + p_{i}/2$ where $p_{i}$ is irrational. These points form a countable set and hence we may have a countable set of points in $[0, 1]$ which do not lie in the interior of some interval $[q_{i} - p_{i}/2, q_{i} + p_{i}/2]$. We can cover these countable points with a sequence of open intervals whose total length does not exceed $\epsilon$. This new sequence of intervals together with $(q_{i} - p_{i}/2, q_{i} + p_{i}/2)$ (note the use of open intervals here) now forms an open cover for $[0, 1]$ and the total length of intervals in this cover is less than $2\epsilon$. Now we can apply the Heine Borel Theorem to reduce this to a finite cover for $[0, 1]$ and the total length of intervals included in the finite cover is less than $2\epsilon$. We get an obvious contradiction if $2\epsilon < 1$. It follows that the assumption $(1)$ is wrong. And that is the fault with OP's proof.

I urge readers to have a look at a similar discussion where the OP is audacious enough to show that measure of whole real line is $0$.

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Answer 3

By assuming that $\sum_{i}p_i<\epsilon$ you're effectively saying that the sum of the lengths of the interval $[-p_i/2,p_i/2]$ is $\epsilon$. In particular, for small $\epsilon$ this length is less then $1$, and so your intervals will never cover all of $[0,1]$.