Egorov's Theorem

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 $\frac{1}{n}$.

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.

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.