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I am trying to answer the second part in this question:

Suppose $f$ and $g$ are continuous functions on $[a,b].$ Show that if $f = g$ a.e. on $[a,b],$ then, in fact, $f = g$ on $[a,b].$ Is a similar assertion true if $[a,b]$ is replaced by a general measurable set $E$?

Here is a counter example I found here Prob. 1, Chap. 3, in Royden's REAL ANALYSIS: If continuous functions $f$ and $g$ agree a.e. on $[a,b]$, then $f=g$ on $[a,b]$ :

The set $[0,1]\cup\{2\}$ is measurable. Let $f(x) = x$ for $x$ in this set, and let $g(x)= x$ for $x\in[0,1]$ and $g(2)=3.$ Then $f$ and $g$ agree almost everywhere on their domain, and both are continuous on that domain, but they don't agree everywhere.

My question is:

I know that the main idea that we used in the proof of the first part was "every nonempty open set in $E = [a,b]$ has positive measure", but I do not understand how by changing our $E$ to $[0,1] \cup \{2\}$ we broke this idea , could anyone explain this to me please?

Brain
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  • One difference: ${2}$ is an open subset of $E$ of measure zero. – BrianO Oct 27 '21 at 05:11
  • why it is an open subset of $E$ and not closed?@BrianO – Brain Oct 27 '21 at 05:12
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    The singleton is open with respect to subspace topology that $E$ inherits from the real line (because for example ${2}=(1,10) \cap E$ is an open subset of the real line intersected with $E$), but clearly the singleton has zero Lebesgue measure. – peek-a-boo Oct 27 '21 at 05:14
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    It's both open and closed. Open because it's an intersection of $E$ with an open subset $U$ of $\mathbb{R}$, for example $U = (3/2, 5/2)$. Thus, another difference: $E$ is disconnected, unlike any interval. – BrianO Oct 27 '21 at 05:15
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    Also, you seem to be forgetting the important saying in topology "sets are not doors" – peek-a-boo Oct 27 '21 at 05:17
  • As far as I understand singletons are closed in $\mathbb R$ with respect to the standard topology on $\mathbb R.$ So why we needed the subspace topology in our example?@peek-a-boo – Brain Oct 27 '21 at 05:20
  • As far as I understand singletons are closed in $\mathbb R$ with respect to the standard topology on $\mathbb R.$ So why we needed the subspace topology in our example?@BrianO – Brain Oct 27 '21 at 05:21
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    The topology on $E$ is the standard topology, ... on $E$ :) Singletons are closed in $E$ too. It's a counterexample! – an example of a less nice domain for which the conclusion of the theorem fails. – BrianO Oct 27 '21 at 05:22
  • @peek-a-boo I have never heard this statement (sets are not doors) before :). – Brain Oct 27 '21 at 15:02

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The main idea used in the proof was that if $f = g$ a.e. on $[a, b]$, then they agree on a dense subset of $[a, b]$, and hence by continuity agree on $[a, b]$. With your example, $f$ and $g$ do not agree on a dense subset of $[0, 1] \cup \{2\}$ since not every open ball around $\{2\}$ contains a point at which $f$ and $g$ agree.

Mason
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