Egorov's Theorem

Let $f_n(x) = 1_{[1/(n+1),\,1/n]}(x)$ ( $0 \le x \le 1$ ). For $0 < \delta \le 1$ , show that the bad set $E_{n,\delta} = \{x : |f_k(x)| \ge \delta \text{ for some } k \ge n\}$ equals $(0, 1/n]$ , and prove completely that for every $\varepsilon > 0$ , choosing $n > 1/\varepsilon$ gives $m(E_{n,\delta}) < \varepsilon$ while $f_k \to 0$ uniformly on $[1/n, 1]$ — Egorov's theorem for this sequence.

E_{n,\delta} = \bigcup_{k \ge n}\left[\frac{1}{k+1}, \frac{1}{k}\right] = \left(0, \frac{1}{n}\right], \quad m(E_{n,\delta}) = \frac{1}{n}

\sup_{x \in [1/n,\,1]} |f_k(x)| = 0 \quad (k \ge n+1)

\delta = 0.5

,

n = 6

: the bad set

(0, \frac16]

has measure

\frac16 \approx 0.1667

. On the good set

[\frac16, 1]

,

f_k \equiv 0

for

k \ge 7

— uniform convergence you can see.

Show proof

Since $0 < \delta \le 1$ , $|f_k(x)| \ge \delta$ is equivalent to $f_k(x) = 1$ , i.e. to $x \in [\frac{1}{k+1}, \frac{1}{k}]$ . Hence $E_{n,\delta} = \bigcup_{k \ge n} [\frac{1}{k+1}, \frac{1}{k}]$ . If $x$ lies in this union, then for some $k \ge n$ , $0 < x \le \frac{1}{k} \le \frac{1}{n}$ , so $x \in (0, \frac{1}{n}]$ . Conversely, if $0 < x \le \frac{1}{n}$ , the intervals $(\frac{1}{k+1}, \frac{1}{k}]$ partition $(0, 1]$ , so exactly one $k$ contains $x$ ; from $\frac{1}{k+1} < x \le \frac{1}{n}$ we get $k + 1 > n$ , i.e. $k \ge n$ . (At $x = 0$ every $f_k$ is 0, so 0 is not in the bad set.) Thus $E_{n,\delta} = (0, \frac{1}{n}]$ with measure $\frac{1}{n}$ .
Given $\varepsilon > 0$ , pick $n > 1/\varepsilon$ (e.g. $n = \lfloor 1/\varepsilon \rfloor + 1$ ). Then $m(E_{n,\delta}) = 1/n < \varepsilon$ — the bad set can be made arbitrarily small.
Now check uniform convergence on $F = [\frac{1}{n}, 1]$ . If $k \ge n+1$ and $x \ge \frac{1}{n}$ , then $x \ge \frac{1}{n} > \frac{1}{n+1} \ge \frac{1}{k}$ , so $x$ lies strictly above the right endpoint of the bump $[\frac{1}{k+1}, \frac{1}{k}]$ — hence $f_k(x) = 0$ . So $\sup_{x \in F} |f_k(x)| = 0$ for every $k \ge n+1$ , and $f_k \to 0$ uniformly on $F$ (the single term $f_n$ equals 1 at the point $\frac{1}{n}$ , but one term never affects the tail).
In summary: $f_n \to 0$ pointwise on all of $[0, 1]$ (but not uniformly, since $\sup f_n = 1$ on $[0,1]$ ), yet deleting the single bad set $(0, \frac{1}{n}]$ of measure below $\varepsilon$ leaves uniform convergence. That is Egorov's theorem for this sequence, visible in the scene as the red region shrinking at rate $\frac{1}{n}$ .

Fix $\delta \le 1$ and let $B_n = E_{n,\delta}$ be the bad sets. Show they form a decreasing sequence ( $B_{n+1} \subseteq B_n$ ) with $m(B_n) = 1/n \to 0$ , by directly proving continuity of measure from above ( $m(A_n) \downarrow m(\bigcap_n A_n)$ ) for finite measure.

B_{n+1} = \left(0, \frac{1}{n+1}\right] \subseteq \left(0, \frac{1}{n}\right] = B_n

m(A_1) = m\!\left(\bigcap_n A_n\right) + \sum_{k \ge 1}\big(m(A_k) - m(A_{k+1})\big)

\delta = 0.5

,

n = 12

:

B_{12} = (0, \frac{1}{12}]

, measure

\frac{1}{12} \approx 0.0833

— nested entirely inside

B_6 = (0, \frac16]

.

Show proof

From problem 1, $B_n = (0, \frac{1}{n}]$ , so $B_{n+1} = (0, \frac{1}{n+1}] \subseteq (0, \frac{1}{n}] = B_n$ is immediate (it also follows from the definition: the union over $k \ge n+1$ sits inside the union over $k \ge n$ ). Hence $\{B_n\}$ is decreasing with $m(B_1) = 1 < \infty$ .
Now prove continuity from above directly. Let $A_n$ be decreasing with $m(A_1) < \infty$ . Decompose $A_1$ into disjoint pieces: $A_1 = \left(\bigcap_n A_n\right) \sqcup \bigsqcup_{k \ge 1} (A_k \setminus A_{k+1})$ . By countable additivity, $m(A_1) = m(\bigcap_n A_n) + \sum_{k \ge 1} \big(m(A_k) - m(A_{k+1})\big)$ (the difference of measures needs $m(A_{k+1}) \le m(A_1) < \infty$ ). The partial sums telescope: $\sum_{k=1}^{N} \big(m(A_k) - m(A_{k+1})\big) = m(A_1) - m(A_{N+1})$ , and bounded above by $m(A_1)$ they converge. Hence $m(A_1) = m(\bigcap_n A_n) + m(A_1) - \lim_{N \to \infty} m(A_N)$ , i.e. $\lim_{N \to \infty} m(A_N) = m(\bigcap_n A_n)$ — finiteness $m(A_1) < \infty$ is what lets us cancel $m(A_1)$ from both sides.
Apply this to $B_n$ : $\bigcap_n B_n = \varnothing$ . Indeed, if $x$ belonged to every $B_n$ , then $0 < x \le \frac{1}{n}$ for all $n$ , forcing $x = 0$ — contradicting $x > 0$ . Hence $m(B_n) \to m(\varnothing) = 0$ , matching the direct computation $m(B_n) = 1/n$ . The bad sets can shrink to empty intersection precisely because the total measure is finite — and that is the heart of Egorov's theorem.

Let $f_n = 1_{[n,\,n+1]}$ (functions on $\mathbb{R}$ ). Prove that for every set $F \subseteq \mathbb{R}$ on which $f_n \to 0$ uniformly, $m(\mathbb{R} \setminus F) = \infty$ — the counterexample showing why the finite-measure hypothesis is essential.

\sup_{x \in F}|f_n(x)| < \tfrac12 \;(n \ge N) \;\Rightarrow\; F \cap [n, n+1] = \varnothing

[N, \infty) \subseteq \mathbb{R}\setminus F \;\Rightarrow\; m(\mathbb{R}\setminus F) = \infty

This problem is about the "escaping spike"

1_{[n,\,n+1]}

on

\mathbb{R}

, not the scene's sequence — the preset is the scene default

\delta = 0.5

,

n = 6

.

Show proof

First, fixing any point $x$ , $f_n(x) = 0$ for $n > x$ , so $f_n \to 0$ pointwise on all of $\mathbb{R}$ . But no finite-measure deletion can restore uniform convergence.
Suppose $f_n \to 0$ uniformly on $F \subseteq \mathbb{R}$ . Then some $N$ has $\sup_{x \in F} |f_n(x)| < \frac{1}{2}$ for all $n \ge N$ . But $f_n$ takes only the values 0 and 1, so a supremum below $\frac{1}{2}$ forces $f_n \equiv 0$ on $F$ — i.e. $F \cap [n, n+1] = \varnothing$ for every $n \ge N$ (otherwise $f_n = 1$ at that point and the supremum would be 1).
Hence $[N, \infty) = \bigcup_{n \ge N} [n, n+1] \subseteq \mathbb{R} \setminus F$ , and $m(\mathbb{R} \setminus F) \ge m([N, \infty)) = \infty$ . Any set with uniform convergence must miss all but finitely many spikes, so the discarded set always has infinite measure — which is why Egorov's theorem cannot drop the hypothesis $m(X) < \infty$ . Contrast the scene's sequence: there the bumps pile up at the single point 0, so deleting a neighborhood of that point suffices.

Examples & Exercises