Set of Discontinuities of Baire Function is Meager

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Theorem

Let $X \subseteq \R$.

Let $f : X \to \R$ be a Baire function.

Let $D$ be the set of points for which $f$ is discontinuous.

Let $d$ be the Euclidean metric on $\R$.

Then $D$ is meager in the metric space $\struct {X, d}$.

Proof

Since $f$ is a Baire function, there exists a sequence $\sequence {f_n}$ of continuous functions such that:

$\sequence {f_n}$ converges pointwise to $f$.

For each $\epsilon > 0$, define:

$\map F \epsilon = \set {x \in X : \map {\omega_f} x > \epsilon}$

where $\map {\omega_f} x$ denotes the oscillation of $f$ at $x$.

From Real Function is Continuous at Point iff Oscillation is Zero:

$f$ is discontinuous at $x$ if and only if $\map {\omega_f} x > 0$.

Therefore:

$\ds D = \bigcup_{i \mathop = 1}^\infty \map F {\frac 1 n}$

We aim to show that $\map F \epsilon$ is nowhere dense in $\struct {X, d}$ for each $\epsilon > 0$.

Let $I$ be an interval.

We aim to show that there exists a subinterval $J$ of $I$ that is disjoint from $\map F \epsilon$.

From this, it will follow that $\map F \epsilon$ is nowhere dense from Open Ball Characterization of Denseness.

For each natural number $n$, let:

\(\ds E_n\)

\(=\)

\(\ds \bigcap_{i \mathop = n}^\infty \bigcap_{j \mathop = n}^\infty \set {x \in I : \size {\map {f_i} x - \map {f_j} x} \le \dfrac \epsilon 4}\)

\(\ds \)

\(=\)

\(\ds \set {x \in I : \size {\map {f_i} x - \map {f_j} x} \le \dfrac \epsilon 4 \, \text { for all } \, i, j \ge n}\)

We can show that each $E_n$ is closed.

For each natural number $i, j$ define $g_{i, j} : X \to \R$ to have:

$\map {g_{i, j} } x = \size {\map {f_i} x - \map {f_j} x}$

From:

Combined Sum Rule for Continuous Real Functions

Absolute Value of Continuous Real Function is Continuous

We have:

$g_{i, j}$ is continuous for each $i, j$.

From Continuity Defined from Closed Sets, we have:

$\map {g^{-1}_{i, j} } {\closedint 0 {\dfrac \epsilon 4} }$ is closed for each $i, j$.

Note that we can write:

$\ds E_n = \bigcap_{i \mathop = n}^\infty \bigcap_{j \mathop = n}^\infty \map {g^{-1}_{i, j} } {\closedint 0 {\dfrac \epsilon 4} }$

This is an intersection of closed sets, so is itself closed by Intersection of Closed Sets is Closed.

Let:

$\ds E = \bigcup_{i \mathop = 1}^\infty E_n$

Clearly:

$E \subseteq I$

Since $\sequence {f_n}$ converges pointwise, the sequence $\sequence {\map {f_n} x}$ is Cauchy, from Convergent Sequence is Cauchy Sequence.

So, for each $x \in I$ there exists an $N$ such that:

$\size {\map {f_i} x - \map {f_j} x} \le \dfrac \epsilon 4$

for all $i, j \ge N$.

That is:

$x \in E_N \subseteq E$

So:

$I \subseteq E$

giving:

$E = I$

From Real Interval is Non-Meager:

$I$ is not meager in the metric space $\struct {X, d}$.

So:

at least one $E_n$ is not nowhere dense in $\struct {X, d}$.

Fix this $n$.

Since $E_n$ is not nowhere dense, there exists some $J \subseteq E_n$ such that:

$E_n \cap J$ is dense in $J$.

That is:

$\paren {E_n \cap J}^- = J$

From Closure of Intersection is Subset of Intersection of Closures, we have:

$\paren {E_n \cap J}^- \subseteq E_n^- \cap J^-$

From Set is Closed iff Equals Topological Closure:

$E_n^- = E_n$

so, we have:

$J \subseteq E_n \cap J^-$

We can therefore see that:

$J \subseteq E_n$

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We therefore have, for $x \in J$:

$\size {\map {f_i} x - \map {f_j} x} \le \dfrac \epsilon 4$

for all $i, j \ge n$.

Taking $j = n$ and $i \to \infty$, we obtain:

$\size {\map f x - \map {f_n} x} \le \dfrac \epsilon 4$

We now look at the difference:

$\size {\map f y - \map f z}$

We have, by the Triangle Inequality:

$\size {\map f y - \map f z} \le \size {\map f y - \map {f_n} y} + \size {\map {f_n} y - \map {f_n} z} + \size {\map {f_n} z - \map f z}$

Since each $f_n$ is continuous on $X$:

$f_n$ is uniformly continuous on compact subsets of $X$

by Continuous Function on Compact Space is Uniformly Continuous.

So, there exists a closed interval $K \subseteq J$ such that:

$\size {\map {f_n} z - \map {f_n} y} \le \dfrac \epsilon 4$

for all $y, z \in K$.

That is:

$\size {\map f y - \map f z} \le \dfrac {3 \epsilon} 4$

for all $y, z \in K$.

So, for any $t \in K$, we have:

$\map {\omega_f} t \le \dfrac {3 \epsilon} 4$

So, $K$ is disjoint from $\map F \epsilon$.

So each $\map F \epsilon$ is nowhere dense in $\struct {X, d}$.

Since $D$ is the countable union of nowhere dense sets in $\struct {X, d}$, it is meager in $\struct {X, d}$.

$\blacksquare$