On Lusin's Theorem we met Littlewood's three principles and proved the second one: every measurable function is nearly continuous. This page is the third.
Littlewood's third principle says that every convergent sequence of measurable functions is nearly uniformly convergent. Here's what that means, and why "nearly" is doing real work. Take the most famous sequence that converges at every point but not uniformly: $x^n$ on $[0, 1]$. At every $x < 1$ it dies off to $0$, and at $x = 1$ it sits at $1$. But the points just left of $1$ take forever to die, so no single $n$ makes the whole graph small at once.
Now cut away a sliver $(1 - \delta, 1]$ at the right end. On what's left, the worst point is $1 - \delta$, and $(1 - \delta)^n \to 0$. The bad behavior lived on a set we can make as small as we like.
Egoroff's theorem says this always happens on a finite measure space: whatever sequence you have, if it converges almost everywhere, you can throw away a set of measure less than $\varepsilon$ and get uniform convergence on the rest. That's the third principle, made precise.
First, the kind of convergence we're comparing against. Pointwise convergence lets every point pick its own $N$. Uniform convergence demands one $N$ that works for every point at once.
Let $f_n, f\colon X \to \mathbb{C}$ and $E \subseteq X$. The sequence $(f_n)$ converges uniformly to $f$ on $E$ if for every $\eta > 0$ there is an $N$ such that $|f_n(x) - f(x)| < \eta$ for all $n \ge N$ and all $x \in E$.
Equivalently, $\sup_{x \in E} |f_n(x) - f(x)| \to 0$: the worst point on $E$ has to behave. That's the number we'll keep computing. (We use $\eta$ here because $\varepsilon$ is about to be busy measuring sets.)
On The Metric Topology, the uniform limit theorem said a uniform limit of continuous functions is continuous. That's exactly why $x^n$ can't converge uniformly on $[0, 1]$: its limit jumps at $1$. We'll lean on the same theorem twice more on this page.
Now the measure-theory version. We're allowed to ignore a small set, measured by $\mu$.
Let $(X, \mathcal{M}, \mu)$ be a measure space, and let $f_n, f\colon X \to \mathbb{C}$ be measurable. The sequence $(f_n)$ converges almost uniformly to $f$ if for every $\varepsilon > 0$ there is a measurable set $E \subseteq X$ with $\mu(E^c) < \varepsilon$ such that $f_n \to f$ uniformly on $E$.
The set $E$ depends on $\varepsilon$: a smaller $\varepsilon$ usually means a bigger $E$, and convergence there may take longer to kick in. Here's the opening example, checked against the definition.
On $[0, 1]$ with Lebesgue measure $m$, show that $x^n$ converges almost uniformly to $f = \chi_{\{1\}}$, but not uniformly on $[0, 1]$.
Pointwise: for $0 \le x < 1$, $x^n \to 0 = f(x)$, and at $x = 1$, $1^n = 1 = f(1)$.
Almost uniformly: given $\varepsilon > 0$, let $\delta = \min\{\frac{\varepsilon}{2}, \frac{1}{2}\}$ and $E = [0, 1 - \delta]$. Then $m(E^c) = \delta < \varepsilon$, and on $E$ the limit is $0$ and $x^n \le (1 - \delta)^n$, so $$\sup_{x \in E} |x^n - f(x)| = (1 - \delta)^n \to 0.$$
Not uniformly on $[0, 1]$: the point $x_n = (\frac{1}{2})^{1/n}$ is less than $1$ and has $x_n^n = \frac{1}{2}$ while $f(x_n) = 0$, so $\sup_{[0, 1]} |x^n - f| \ge \frac{1}{2}$ for every $n$.
So almost uniform sits strictly between uniform and pointwise. Watch out for the name, though, because it invites a misreading.
"Almost uniformly" does not mean "uniformly almost everywhere", that is, uniformly off a set of measure $0$. Almost uniform lets you remove a set of small positive measure, and a different one for each $\varepsilon$. For $x^n$ on $[0, 1]$, no null set can be removed to make the convergence uniform. Some books call almost uniform convergence "nearly uniform"; same idea.
Why can't a null set do the job? Removing a set of measure $0$ can't remove an interval, so points close to $1$ always survive, and $x^n$ is still near $1$ at some of them.
Prove it. The animation is the whole idea; the exercise is making "some point survives" precise.
On $[0, 1]$ with Lebesgue measure $m$, let $E \subseteq [0, 1]$ be measurable with $m(E^c) = 0$. Show that $x^n$ does not converge uniformly to $\chi_{\{1\}}$ on $E$.
Proof. Fix $n$. The interval $I_n = ((\frac{1}{2})^{1/n}, 1)$ has positive length, and $m(E^c) = 0$, so $I_n \not\subseteq E^c$: there is a point $x \in E \cap I_n$. At that point $x < 1$, so the limit is $0$, while $x^n > \frac{1}{2}$. Hence $$\sup_{x \in E} |x^n - \chi_{\{1\}}(x)| > \tfrac{1}{2} \quad \text{for every } n,$$ and the sup does not tend to $0$. $\blacksquare$
One direction of the comparison is free, on every measure space: almost uniform convergence is strong enough to give convergence almost everywhere. Try it before reading the answer.
Let $(X, \mathcal{M}, \mu)$ be any measure space. Show that if $f_n \to f$ almost uniformly, then $f_n \to f$ $\mu$-almost everywhere.
Proof. For each $m \ge 1$, pick a measurable $E_m$ with $\mu(E_m^c) < \frac{1}{m}$ on which $f_n \to f$ uniformly. Uniform convergence on $E_m$ gives pointwise convergence there, so $f_n(x) \to f(x)$ for every $x$ in $G = \bigcup_m E_m$, which is measurable.
Its complement $G^c = \bigcap_m E_m^c$ lies inside every $E_m^c$, so $\mu(G^c) \le \mu(E_m^c) < \frac{1}{m}$ for every $m$. Hence $\mu(G^c) = 0$, and $f_n \to f$ $\mu$-almost everywhere. $\blacksquare$
So almost uniform implies almost everywhere, always. Egoroff's theorem is the converse, and the converse is the part that needs a hypothesis.
The proof sorts points by how well behaved they are. For a tolerance $\frac{1}{k}$ and a starting index $n$, collect the points where the tail of the sequence, from $n$ on, never wobbles by $\frac{1}{k}$ or more: $$S_{n,k} = \bigcap_{i, j \ge n} \left\{x : |f_i(x) - f_j(x)| < \tfrac{1}{k}\right\}.$$ Before the general proof, compute these sets for our favorite sequence.
For $f_n(x) = x^n$ on $[0, 1]$, find $S_{n,k}$ and $m(S_{n,k}^c)$ for $k \ge 2$.
At $x = 1$ every $f_i(x) = 1$, so $1 \in S_{n,k}$. For $0 \le x < 1$ and $n \le i < j$, $|x^i - x^j| = x^i - x^j < x^i \le x^n$, so if $x^n \le \frac{1}{k}$, every difference is less than $\frac{1}{k}$. If instead $x^n > \frac{1}{k}$, then $|x^n - x^j| \to x^n > \frac{1}{k}$ as $j \to \infty$, so some difference is too big. Hence $$S_{n,k} = \{x \in [0, 1) : x^n \le \tfrac{1}{k}\} \cup \{1\} = [0, k^{-1/n}] \cup \{1\}, \qquad m(S_{n,k}^c) = 1 - k^{-1/n}.$$
As $n \to \infty$, $k^{-1/n} \to 1$, so the bad set $S_{n,k}^c = (k^{-1/n}, 1)$ shrinks to nothing.
That's the pattern: for each fixed $k$, the sets $S_{n,k}$ grow with $n$, and they fill up the space, because at each point the sequence converges, so its tail eventually stops wobbling. Here they are growing for $k = 4$.
On Measures, continuity from below said that if sets grow, $A_1 \subseteq A_2 \subseteq \cdots$, then $\mu(A_n) \to \mu(\bigcup_n A_n)$. Growing sets that fill $X$ have measures tending to $\mu(X)$.
Growing sets that fill $X$, so continuity from below gives $\mu(S_{n,k}) \to \mu(X)$. Here's where the finite measure comes in: only when $\mu(X) < \infty$ can we subtract and conclude that the bad set's measure $\mu(S_{n,k}^c)$ goes to $0$. Then pick, for each $k$, an $n_k$ whose bad set costs at most $\varepsilon 2^{-k}$, and keep the points that are good at every level.
Let $(X, \mathcal{M}, \mu)$ be a measure space with $\mu(X) < \infty$, let $f_n\colon X \to \mathbb{C}$ be measurable, and suppose $f_n(x) \to f(x) \in \mathbb{C}$ for $\mu$-almost every $x$. Then for every $\varepsilon > 0$ there is a measurable $E \subseteq X$ with $\mu(E^c) < \varepsilon$ such that $f_n \to f$ uniformly on $E$. That is, $f_n \to f$ almost uniformly.
Proof. By hypothesis there is a null set $N \in \mathcal{M}$ such that $f_n(x) \to f(x)$ for every $x \in Y = X \setminus N$. For $n, k \ge 1$ let $$S_{n,k} = \bigcap_{i, j \ge n} \left\{x \in Y : |f_i(x) - f_j(x)| < \tfrac{1}{k}\right\}.$$ Each $|f_i - f_j|$ is measurable, so each set in the intersection is measurable, and a countable intersection of measurable sets is measurable.
The sets fill $Y$. Fix $k$. If $n \le n'$, the intersection defining $S_{n',k}$ has fewer conditions, so $S_{1,k} \subseteq S_{2,k} \subseteq \cdots$. For $x \in Y$, the sequence $(f_n(x))$ converges, so it is Cauchy: there is $n$ with $|f_i(x) - f_j(x)| < \frac{1}{k}$ for all $i, j \ge n$, which says $x \in S_{n,k}$. So $\bigcup_n S_{n,k} = Y$.
The bad sets shrink. By continuity from below, $\mu(S_{n,k}) \to \mu(Y) = \mu(X)$ as $n \to \infty$. Since $\mu(X) < \infty$, we may subtract: $\mu(X \setminus S_{n,k}) = \mu(X) - \mu(S_{n,k}) \to 0$. Given $\varepsilon > 0$, choose for each $k$ an index $n_k$ with $\mu(X \setminus S_{n_k,k}) < \varepsilon 2^{-k}$.
The set $E$. Let $E = \bigcap_{k} S_{n_k,k}$, which is measurable. Then $E^c = \bigcup_k (X \setminus S_{n_k,k})$, so by countable subadditivity $$\mu(E^c) \le \sum_{k=1}^{\infty} \mu(X \setminus S_{n_k,k}) < \sum_{k=1}^{\infty} \varepsilon 2^{-k} = \varepsilon.$$ (The inequality is strict because every term is strictly smaller and the series on the right converges.)
Uniform convergence on $E$. Let $\eta > 0$ and pick $k$ with $\frac{1}{k} < \eta$. Every $x \in E$ lies in $S_{n_k,k}$, so $|f_i(x) - f_j(x)| < \frac{1}{k}$ for all $i, j \ge n_k$. Since $x \in Y$, letting $j \to \infty$ gives $|f_i(x) - f(x)| \le \frac{1}{k} < \eta$ for all $i \ge n_k$. The index $n_k$ does not depend on $x$, so $f_n \to f$ uniformly on $E$. $\blacksquare$
Egoroff: uniform convergence, terms and conditions apply.
The heart of the proof is the error budget: the bad set at level $k$ gets $\varepsilon 2^{-k}$ of room, and the whole bill comes to less than $\varepsilon$, however many levels there are.
Notice what the theorem doesn't say: it doesn't tell you how long you wait. Cut less, and you wait longer. Feel the trade-off with three different sequences.
The trouble spot can sit at either end, and the spikes $n\chi_{(0, 1/n)}$ show something strange: the waiting time doesn't depend on $\varepsilon$ at all, because the spike either sits on $E$ (too big for any $\varepsilon < 1$) or has left it.
Can I just throw away all of $[0, 1]$? Uniform convergence on the empty set is very easy. ...Measure $1$ isn't less than $\varepsilon$. Fine.
Your turn to compute one. Find where on $E$ the error is worst, write it as a function of $n$, and solve.
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Now run the proof yourself on $x^n$ with $\varepsilon = \frac{1}{2}$. At each level $k$ you pick the index $n_k$, and its bad set has to fit its share of the budget. (Level $k = 1$ is free here: every difference $|x^i - x^j|$ on $[0, 1]$ is already less than $1$.)
For $x^n$ the bad sets are nested intervals at the right end, so $E^c$ is just the biggest one. The proof can't count on that, which is why it pays for every level separately. Check that you can put the proof back together.
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The proof used $\mu(X) < \infty$ exactly once, to turn $\mu(S_{n,k}) \to \mu(X)$ into $\mu(S_{n,k}^c) \to 0$. Without it, the theorem really does fail, and the counterexample is a bump that runs away. On $[0, \infty)$ with Lebesgue measure, $\chi_{[n, \infty)} \to 0$ at every single point: each $x$ gets left behind once $n > x$. But the bump itself never shrinks.
A set of finite measure can't swallow a tail $[n, \infty)$, so whatever you remove, the bump keeps finding points of $E$. Try to beat it anyway, and then change the measure.
With $e^{-x}\,dx$ the whole half-line has measure $1$, so a tail $[a, \infty)$ costs only $e^{-a}$, and cutting it away is cheap. That's Egoroff working, because the new measure is finite. Now write down why Lebesgue measure loses, every time.
On $[0, \infty)$ with Lebesgue measure $m$, let $f_n = \chi_{[n, \infty)}$. Show that $f_n \to 0$ at every point, but there is no measurable $E$ with $m(E^c) < \infty$ on which $f_n \to 0$ uniformly. (In particular, $f_n$ doesn't converge almost uniformly.)
Proof. For each $x \ge 0$, once $n > x$ we have $x \notin [n, \infty)$, so $f_n(x) = 0$. Hence $f_n(x) \to 0$ at every point.
Let $E$ be measurable with $m(E^c) < \infty$, and fix $n$. Since $m([n, \infty)) = \infty > m(E^c)$, the tail $[n, \infty)$ is not contained in $E^c$, so there is $x \in E \cap [n, \infty)$, where $f_n(x) = 1$. So $\sup_{x \in E} |f_n(x) - 0| = 1$ for every $n$, and the convergence on $E$ is not uniform. $\blacksquare$
Careful with how this counterexample gets described. Every $f_n(x)$ does converge; the failure is that $\sup_E f_n = 1$ forever. And the culprit isn't σ-finiteness: $[0, \infty)$ is σ-finite. It's that $\mu(X) = \infty$.
The runaway bump is one way to fail. Here's another, where nothing runs away at all: the function just grows slowly everywhere at once.
On $\mathbb{R}$ with Lebesgue measure $m$, does $f_n(x) = \frac{x}{n}$ converge almost uniformly?
Pointwise, yes: for each fixed $x$, $\frac{x}{n} \to 0$.
Almost uniformly, no. Suppose $m(E^c) < 1$. Then $E^c$ can't contain any interval $[a, a + 1]$, so $E$ meets every such interval, and $E$ is unbounded. For each $n$ and each $R > 0$ there is $x \in E$ with $|x| > R$, so $\sup_{x \in E} \frac{|x|}{n} = \infty$. The sup never even becomes finite, let alone small.
So on an infinite measure space, convergence can fail to be almost uniform even when the functions are as tame as straight lines. But there's a hypothesis that rescues Egoroff without a finite measure: one integrable function sitting on top of the whole sequence.
On Integrable Functions, Markov's inequality said $\mu(\{|g| > t\}) \le \frac{1}{t}\int_X |g|\,d\mu$. An integrable function can only be large on a set of finite measure.
Let $(X, \mathcal{M}, \mu)$ be any measure space, let $f_n\colon X \to \mathbb{C}$ be measurable with $f_n \to f$ $\mu$-a.e., and suppose there is $g \in L^1(\mu)$ with $|f_n| \le g$ $\mu$-a.e. for every $n$. Show that $f_n \to f$ almost uniformly.
Proof. Collect the countably many exceptional null sets into one null set $N$, so that on $Y = X \setminus N$ we have $f_n \to f$ and $|f_n| \le g$ for every $n$. Define $S_{n,k}$ as in Egoroff's proof, and let $D_{n,k} = Y \setminus S_{n,k} = \bigcup_{i, j \ge n} \{x \in Y : |f_i(x) - f_j(x)| \ge \frac{1}{k}\}$.
On $Y$, $|f_i - f_j| \le 2g$, so $D_{n,k} \subseteq \{g \ge \frac{1}{2k}\} \subseteq \{g > \frac{1}{4k}\}$, and by Markov's inequality $$\mu(D_{1,k}) \le \mu(\{g > \tfrac{1}{4k}\}) \le 4k \int_X g \, d\mu < \infty.$$ For fixed $k$ the sets $D_{n,k}$ decrease in $n$, and their intersection is empty because the $S_{n,k}$ fill $Y$. Since $\mu(D_{1,k}) < \infty$, continuity from above gives $\mu(D_{n,k}) \to 0$. That is the only place Egoroff's proof used $\mu(X) < \infty$; the rest goes through word for word: choose $n_k$ with $\mu(D_{n_k,k}) < \varepsilon 2^{-k}$, let $E = \bigcap_k S_{n_k,k}$, so $E^c = N \cup \bigcup_k D_{n_k,k}$ has measure less than $\varepsilon$, and $f_n \to f$ uniformly on $E$. $\blacksquare$
That's why $\chi_{[n, \infty)}$ escapes: no integrable function sits on top of all of them, since such a $g$ would have to be at least $1$ on all of $[1, \infty)$. You now have the whole ladder: uniform, almost uniform, almost everywhere. Sort these sequences onto it.
The measure matters as much as the functions: change Lebesgue measure to a point mass and a sequence can jump a whole rung. Name the strongest mode for a few more.
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Egoroff comes back on The Lᵖ Spaces, a page that isn't written yet. A sequence converging in $L^p$ has a subsequence converging almost everywhere, and on a finite measure space Egoroff then upgrades it to almost uniform.
Egoroff has a famous application: a second proof of Lusin's theorem. It goes like this. Approximate $f$ by simple functions, use Egoroff to make the approximation uniform on a big set, and shrink that set to a compact $K$ where everything is continuous. Then $f|_K$ is a uniform limit of continuous functions, so it is continuous on $K$. The last step needs a tool we don't have yet: a continuous function on a compact set $K$ has to be extended to a continuous function on all of $X$.
On Locally Compact Spaces, Urysohn's lemma built $g \in C_c(X)$ with $\chi_K \le g \le \chi_V$ for compact $K$ inside open $V$: an extension of the constant $1$ on $K$. Tietze extends any continuous function on $K$, by stacking Urysohn functions.
On $\mathbb{R}$, extending is just connecting the dots. Start there.
Let $K = [0, 1] \cup [2, 3] \subseteq \mathbb{R}$ and $U = (-1, 4)$, and let $f = 1$ on $[0, 1]$ and $f = -1$ on $[2, 3]$. Find $g \in C_c(\mathbb{R})$ with $g = f$ on $K$ and $\operatorname{supp} g \subseteq U$.
Note that $f$ is continuous on $K$: each piece is constant, and the pieces are apart. Connect the dots, and ramp down to $0$ inside $U$: $$g(x) = \begin{cases} 0, & x \le -\frac{1}{2}, \\ 2x + 1, & -\frac{1}{2} \le x \le 0, \\ 1, & 0 \le x \le 1, \\ 3 - 2x, & 1 \le x \le 2, \\ -1, & 2 \le x \le 3, \\ 2x - 7, & 3 \le x \le \frac{7}{2}, \\ 0, & x \ge \frac{7}{2}. \end{cases}$$ Neighboring formulas agree where they meet ($0$ at $-\frac{1}{2}$, $1$ at $0$ and $1$, $-1$ at $2$ and $3$, $0$ at $\frac{7}{2}$), so $g$ is continuous. It equals $f$ on $K$, and $\operatorname{supp} g = [-\frac{1}{2}, \frac{7}{2}]$ is compact and inside $U$. Also $|g| \le 1 = \sup_K |f|$.
In a general locally compact space there are no "dots" to connect and no straight lines. What we do have is Urysohn's lemma, which can only make one shape: a bump that is $1$ on a compact set. Tietze builds any continuous function out of such bumps by successive approximation.
Let $X$ be a locally compact Hausdorff space, $K \subseteq X$ compact, and $U \subseteq X$ open with $K \subseteq U$. If $f\colon K \to \mathbb{C}$ is continuous, there is $g \in C_c(X)$ with $g|_K = f$ and $\operatorname{supp} g \subseteq U$. If $f$ is real-valued, $g$ can be taken real-valued with $\sup_X |g| \le \sup_K |f|$.
Proof. Reduction to real $f$. Write $f = u + iv$. If $g_u, g_v$ are real extensions of $u, v$ in $C_c(X)$ with supports in $U$, then $g = g_u + i g_v$ is in $C_c(X)$, extends $f$, and $\operatorname{supp} g \subseteq \operatorname{supp} g_u \cup \operatorname{supp} g_v \subseteq U$. So let $f$ be real, and let $M = \sup_K |f|$, which is finite because $f$ is continuous on a compact set. If $M = 0$, take $g = 0$.
One set for every round. By the shrinking theorem, there is an open $W$ with compact closure such that $K \subseteq W \subseteq \overline{W} \subseteq U$. Every function built below will vanish off $W$.
One round. Let $h\colon K \to \mathbb{R}$ be continuous with $|h| \le c$ on $K$, where $c > 0$. We find $\varphi \in C_c(X)$ with $\varphi = 0$ off $W$, $|\varphi| \le \frac{c}{3}$ on $X$, and $|h - \varphi| \le \frac{2c}{3}$ on $K$.
Let $A = \{x \in K : h(x) \le -\frac{c}{3}\}$ and $B = \{x \in K : h(x) \ge \frac{c}{3}\}$. They are closed subsets of $K$ (preimages of closed sets under the continuous $h$), hence compact, and disjoint. Disjoint compact sets in a Hausdorff space lie in disjoint open sets $O_A \supseteq A$ and $O_B \supseteq B$ (the separation fact in the proof of Lusin's theorem). Urysohn's lemma gives $\varphi_B \in C_c(X)$ with $\chi_B \le \varphi_B \le \chi_{O_B \cap W}$ and $\operatorname{supp} \varphi_B \subseteq O_B \cap W$, and likewise $\varphi_A$ for $A$ inside $O_A \cap W$. (If $A$ or $B$ is empty, use the zero function for it.) Let $$\varphi = \tfrac{c}{3}(\varphi_B - \varphi_A).$$ It is continuous with compact support inside $W$. The supports of $\varphi_A$ and $\varphi_B$ are disjoint, and both take values in $[0, 1]$, so $|\varphi| \le \frac{c}{3}$; also $\varphi = \frac{c}{3}$ on $B$ and $\varphi = -\frac{c}{3}$ on $A$.
Check $|h - \varphi| \le \frac{2c}{3}$ on $K$ in three cases. On $B$, $h \in [\frac{c}{3}, c]$ and $\varphi = \frac{c}{3}$, so $h - \varphi \in [0, \frac{2c}{3}]$. On $A$, likewise $h - \varphi \in [-\frac{2c}{3}, 0]$. On the rest of $K$, $|h| < \frac{c}{3}$ and $|\varphi| \le \frac{c}{3}$, so $|h - \varphi| < \frac{2c}{3}$.
The rounds. Apply one round to $h_1 = f$ with $c = M$, giving $\varphi_1$. Apply it to $h_2 = f - \varphi_1$ on $K$, with $c = \frac{2}{3}M$, giving $\varphi_2$, and so on. By induction we get $\varphi_n \in C_c(X)$ vanishing off $W$ with $$|\varphi_n| \le \frac{M}{3}\left(\frac{2}{3}\right)^{n-1} \text{ on } X, \qquad \left|f - (\varphi_1 + \cdots + \varphi_n)\right| \le M\left(\frac{2}{3}\right)^n \text{ on } K.$$
The sum. Since $\sum_n \frac{M}{3}(\frac{2}{3})^{n-1} = M < \infty$, the Weierstrass M-test says $g = \sum_n \varphi_n$ converges uniformly on $X$, so $g$ is continuous, real, and $|g| \le M$. On $K$, the second bound tends to $0$, so $g = f$. Every $\varphi_n$ vanishes off $W$, so $g$ does too, and $\operatorname{supp} g \subseteq \overline{W}$, a compact subset of $U$. Hence $g \in C_c(X)$, and its support is a closed subset of $\overline{W} \subseteq U$. $\blacksquare$
Each round fixes a third of the remaining error, and the error shrinks geometrically. Watch the rounds on the example's set $K$, with a function that isn't piecewise constant.
Tietze shaves a third off the error every round. Very patient. Infinitely patient, technically.
Now draw your own $f$ on $K$ and squeeze the open set $U$. The extension takes whatever room you leave it.
Two hypotheses in the proof are easy to underrate. First, the set $W$. Each $\varphi_n$ has compact support, so why not just add them up and stop worrying? Because infinitely many compact supports can spread out.
Find functions $h_N \in C_c(\mathbb{R})$ that converge uniformly on $\mathbb{R}$ to a function $h \notin C_c(\mathbb{R})$. What does this say about the Tietze proof?
Let $\tau(x) = \max\{0, 1 - |x|\}$, a tent of height $1$ on $[-1, 1]$, and let $$h_N(x) = \sum_{n=1}^{N} 2^{-n} \tau(x - 3n).$$ Each $h_N$ is continuous with support $[2, 3N + 1]$, so $h_N \in C_c(\mathbb{R})$. The terms satisfy $|2^{-n}\tau(x - 3n)| \le 2^{-n}$ and $\sum_n 2^{-n} < \infty$, so by the M-test $h_N \to h = \sum_{n=1}^{\infty} 2^{-n} \tau(x - 3n)$ uniformly, and $h$ is continuous.
The tents sit on the disjoint intervals $[3n - 1, 3n + 1]$, so $h(3n) = 2^{-n} \ne 0$ for every $n$. Then $\{h \neq 0\}$ is unbounded, $\operatorname{supp} h$ is not compact, and $h \notin C_c(\mathbb{R})$. (It does vanish at infinity.)
So "uniform limit of $C_c$ functions" only gives a continuous function, not one with compact support. The Tietze proof keeps every $\varphi_n$ inside one $W$ with compact closure, which traps the support of the sum inside $\overline{W}$. $\blacksquare$
Second, compactness of $K$. On a set that isn't closed, a perfectly good continuous function can refuse to be extended.
Let $K = (0, 1] \subseteq \mathbb{R}$ and $f(x) = \sin(1/x)$ on $K$. Show that $f$ is continuous and bounded on $K$, but there is no continuous $g\colon \mathbb{R} \to \mathbb{R}$ with $g|_K = f$.
Proof. $f$ is a composition of continuous functions on $(0, 1]$, and $|f| \le 1$. Suppose $g$ is continuous on $\mathbb{R}$ and equals $f$ on $K$. Let $x_k = \frac{1}{\pi/2 + 2\pi k}$ and $y_k = \frac{1}{3\pi/2 + 2\pi k}$ for $k \ge 1$. Both sequences lie in $K$ and tend to $0$, with $f(x_k) = 1$ and $f(y_k) = -1$. Continuity of $g$ at $0$ gives $$g(0) = \lim_k g(x_k) = 1 \quad \text{and} \quad g(0) = \lim_k g(y_k) = -1,$$ a contradiction. The set $K$ is not compact (not closed), and that is the hypothesis that fails. $\blacksquare$
The proof's bookkeeping is all powers of $\frac{2}{3}$. Run it on a few numbers.
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The proof on Lusin's Theorem built $g$ in five steps: simple functions with Urysohn bumps, layers of a staircase, a cutoff at height $n$, a split into four parts, and a radial retraction.
Now we have both tools for the second proof. Here's the plan; the details are the Spicy Bonus at the bottom of the page.
Where the first proof built $g$ by hand, this one gets it from two theorems. Here's how the three modes of convergence on this page fit together, and where Egoroff sits among them.
Uniform implies almost uniform, which implies almost everywhere, on every measure space. Neither arrow reverses in general, and Egoroff is exactly the condition, $\mu(X) < \infty$, that sends the second one back.
Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
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Next, the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the types that need their answer choices will say so.
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Last, the Spicy Bonus: the second proof of Lusin's theorem, in full. The famous version of this proof applies Tietze to the simple functions $s_n$ on $K$, but simple functions aren't continuous there, so Step 1 has to make them continuous first.
Assume the standing hypotheses of Lusin's theorem. Let $f\colon X \to \mathbb{C}$ be measurable with $\mu(A) < \infty$, where $A = \{f \neq 0\}$. Using Egoroff's theorem and the Tietze extension theorem (not the five-step proof), show that for every $\varepsilon > 0$ there is $g \in C_c(X)$ with $$\mu(\{f \neq g\}) < \varepsilon \quad \text{and} \quad \sup_{X} |g| \le \sup_{X} |f|.$$
Proof. An open set around $A$. By outer regularity there is an open $U \supseteq A$ with $\mu(U) < \mu(A) + \frac{\varepsilon}{4}$. Since $\mu(A) < \infty$, this says $\mu(U \setminus A) < \frac{\varepsilon}{4}$.
Step 1: a simple function is continuous on a big compact set. Let $s$ be a simple function that vanishes off $A$, and let $\eta > 0$. The sets $L_i = A \cap s^{-1}(\{\alpha_i\})$, one for each value $\alpha_1, \ldots, \alpha_r$ of $s$ (zero included), are measurable, disjoint, and cover $A$. Each has finite measure, so inner regularity gives compact $C_i \subseteq L_i$ with $\mu(L_i \setminus C_i) < \frac{\eta}{r}$. Let $C = C_1 \cup \cdots \cup C_r$, a compact subset of $A$ with $\mu(A \setminus C) < \eta$. The $C_i$ are disjoint compact sets, so they lie in pairwise disjoint open sets $V_i$ (the separation fact from Lusin's proof). Then $C \cap V_i = C_i$, and $s = \alpha_i$ there, so every point of $C$ has a neighborhood on which $s|_C$ is constant. Hence $s|_C$ is continuous.
Step 2: simple functions. Take simple functions $s_n$ with $s_n(x) \to f(x)$ and $|s_n(x)| \le |f(x)|$ at every $x$ (staircases applied to the positive and negative parts of $\operatorname{Re} f$ and $\operatorname{Im} f$). Each $s_n$ vanishes off $A$. Step 1 with $\eta = \varepsilon 2^{-n-2}$ gives compact $C_n \subseteq A$ with $s_n|_{C_n}$ continuous and $\mu(A \setminus C_n) < \varepsilon 2^{-n-2}$.
Step 3: Egoroff. Restricted to $A$, $\mu$ is a finite measure, and $s_n \to f$ at every point of $A$. Egoroff's theorem gives a measurable $E \subseteq A$ with $\mu(A \setminus E) < \frac{\varepsilon}{4}$ on which $s_n \to f$ uniformly.
Step 4: the compact set $K$. Let $F = E \cap \bigcap_n C_n$. It is measurable, and $$\mu(A \setminus F) \le \mu(A \setminus E) + \sum_{n} \mu(A \setminus C_n) < \frac{\varepsilon}{4} + \sum_{n=1}^{\infty} \varepsilon 2^{-n-2} = \frac{\varepsilon}{2}.$$ Since $\mu(F) < \infty$, inner regularity gives compact $K \subseteq F$ with $\mu(F \setminus K) < \frac{\varepsilon}{4}$, so $\mu(A \setminus K) < \frac{3\varepsilon}{4}$. On $K$, each $s_n|_K$ is continuous (it is a restriction of $s_n|_{C_n}$), and $s_n \to f$ uniformly on $K \subseteq E$. By the uniform limit theorem, $f|_K$ is continuous.
Step 5: Tietze. $K$ is compact and $K \subseteq A \subseteq U$, so the Tietze extension theorem gives $g \in C_c(X)$ with $g|_K = f|_K$ and $\operatorname{supp} g \subseteq U$. Off $U$, both $f$ and $g$ are $0$ (since $A \subseteq U$); on $K$ they agree. So $\{f \neq g\} \subseteq U \setminus K$, and $$\mu(\{f \neq g\}) \le \mu(U \setminus A) + \mu(A \setminus K) < \frac{\varepsilon}{4} + \frac{3\varepsilon}{4} = \varepsilon.$$
Step 6: the bound. If $M = \sup_X |f|$ is $\infty$ there is nothing to do. Otherwise replace $g$ by $\varphi \circ g$, where $\varphi(z) = z$ for $|z| \le M$ and $\varphi(z) = Mz/|z|$ for $|z| > M$, exactly as in Step 4 of Lusin's proof: $\varphi \circ g \in C_c(X)$, $|\varphi \circ g| \le M$, and $\varphi(g(x)) = f(x)$ wherever $g(x) = f(x)$, so the bad set doesn't grow. $\blacksquare$