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\((\Omega , \mathcal{F})\) will denote a space together with a sigma-algebra on it (a measurable space).
Definitions
Definition 1 Let \((\Omega , \mathcal{F}), \; (\Omega^\prime , \mathcal{F}^\prime )\) be two arbitrary measurable spaces. A map \(f: \Omega \to \Omega^\prime\) is called \(\mathcal{F} | \mathcal{F}^\prime\)-measurable (or simply measurable, if it is clear from context which sigma-algebras are meant) if the preimages of all elements of \(\mathcal{F}'\) lie in \(\mathcal{F}\):
\[
f^{-1}(B) \in \mathcal{F}, \qquad \forall B \in \mathcal{F}^\prime
\]
This condition can be rewritten as \(f^{-1}(\mathcal{F}^\prime ) \subseteq \mathcal{F}\). The following notation is also used for \(f^{-1}(B)\): \(\left\{ f \in B\right\}\). A measurable function is also written as:
\[
f: (\Omega , \mathcal{F}) \to (\Omega ^\prime , \mathcal{F}^\prime )
\]
Lemma 1 (Measurability criterion) Suppose the sigma-algebra \(\mathcal{F}'\) on the codomain \(\Omega '\) is generated by some family \(\mathcal{A} \subset 2^{\Omega '}\): \(\mathcal{F}' = \sigma (\mathcal{A})\). Then \(f: \Omega \to \Omega '\) is measurable if and only if the preimages of the sets in \(\mathcal{A}\) lie in \(\mathcal{F}\): \(f^{-1}(\mathcal{A}) \subset \mathcal{F}\).
As a rule, we will be interested in numerical measurable maps, i.e., measurable functions. In this case \((\Omega^\prime , \mathcal{F}^\prime ) = (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\). It is convenient to take the extended real line as the codomain right away, since later on we will need, for example, to prove that the supremum of a collection of measurable functions is itself measurable. The open sets in \(\overline{\mathbb {R}} = [-\infty , +\infty ]\) are taken to be sets of the form
\[
A, \quad A \cup [-\infty , x), \quad A \cup (y, +\infty ], \quad A \cup [-\infty ,x) \cup (y, +\infty ]
\]
where \(A \subset \mathbb {R}\) is open, \(x, y \in \mathbb {R}\). \(\mathscr {B}\left(\overline{\mathbb {R}}\right)\) is the Borel sigma-algebra on \(\overline{\mathbb {R}}\), i.e., the minimal sigma-algebra containing all open subsets of \(\overline{\mathbb {R}}\). It is not hard to prove that
\[
\begin{align} \mathscr {B}\left(\overline{\mathbb {R}}\right) = \mathscr {B}\left(\mathbb {R}\right) & \cup \left\{ B \cup \left\{ +\infty \right\} , B \in \mathscr {B}\left(\mathbb {R}\right)\right\} \cup \\ & \cup \left\{ B \cup \left\{ -\infty \right\} , B \in \mathscr {B}\left(\mathbb {R}\right)\right\} \cup \\ & \cup \left\{ B \cup \left\{ -\infty , +\infty \right\} , B \in \mathscr {B}\left(\mathbb {R}\right)\right\} \end{align}
\]
From the measurability criterion and the properties of the Borel sigma-algebra on \(\mathbb {R}\) the following lemma follows.
Lemma 2 (measurability criterion for a function (a numerical map)) Let \(f: \Omega \to \mathbb {R}\), let \(\mathcal{F}\) be a sigma-algebra on \(\Omega\). Let \(D\) be a dense subset of \(\mathbb {R}\). The following conditions are equivalent:
\(f\) is \(\mathcal{F}\mid \mathscr {B}\left(\mathbb {R}\right)\)-measurable;
\(f^{-1}((x, +\infty ]) \in \mathcal{F}, \quad \forall x \in D\)
\(f^{-1}([x, +\infty ]) \in \mathcal{F}, \quad \forall x \in D\)
\(f^{-1}([-\infty , x)) \in \mathcal{F}, \quad \forall x \in D\)
\(f^{-1}([-\infty , x]) \in \mathcal{F}, \quad \forall x \in D\)
\(f^{-1}(B) \in \mathcal{F} \; \forall B \in \mathscr {B}\left(\mathbb {R}\right)\) and \(f^{-1}(\left\{ -\infty \right\} ), f^{-1}(\left\{ +\infty \right\} ) \in \mathcal{F}\).
Definition 2 A measurable function \(f: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) is called Borel measurable if its domain \((\Omega , \mathcal{F})\) is a Borel subset of \(\mathbb {R}^{k}\) with the Borel sigma-algebra: \(\left(\Omega , \mathcal{F}\right) = \left(B, \mathscr {B}\left(B\right)\right)\), \(B \in \mathscr {B}\left(\mathbb {R}^{k}\right)\).
A measurable function \(f: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) is called Lebesgue measurable if its domain is a Lebesgue measurable subset of \(\mathbb {R}^{k}\) with the sigma-algebra of its Lebesgue measurable subsets, i.e., \(\Omega = A \in \mathcal{M}(\lambda_k^*) \subset 2^{\mathbb {R}^{k}}\), \(\mathcal{F} = A \cap \mathcal{M}(\lambda^*_k)\).
Prove that the function \(f = \; \mathbb {1}_{A}\) is \(\mathcal{F}|\mathscr {B}\left(\overline{\mathbb {R}}\right)\)-measurable if and only if \(A \in \mathcal{F}\).
Let \(B \in \mathscr {B}\left(\mathbb {R}^{k}\right)\) be an arbitrary Borel subset of \(\mathbb {R}^{k}\). Prove that a continuous function \(f: B \to \overline{\mathbb {R}}\) is Borel measurable.
A function is continuous (i.e., continuous at every point of its whole domain) if and only if the preimage of every open subset of the codomain is open in the domain.
Construct on \([0,1]\) a Lebesgue measurable function that is discontinuous at every point.
Let \(A_1, \dots , A_n\) be a partition of \(\Omega\) and let \(\mathcal{F} = \sigma (\left\{ A_1, \dots , A_n\right\} )\). Characterize all measurable functions from \((\Omega , \mathcal{F})\) in this case.
Let \(f: [0,1] \to \overline{\mathbb {R}}\) be a function monotone on \([0,1]\). Prove that \(f\) is Borel measurable.
Without loss of generality, let \(f\) be increasing. β¦ to be completed
Prove that \(\mathscr {B}\left(\mathbb {R}\right) \subsetneq \mathcal{M}(\lambda )\). Namely, construct a Lebesgue measurable set that is not Borel.
Give an example of a continuous function \(f: [0,1] \to [0,1]\) and a set \(C \in \mathcal{M}(\lambda^*)\) such that \(C \subset [0,1], \lambda (C) = 0\), but \(\lambda (f(C)) > 0\) (for this, of course, we must require that \(f(C) \in \mathcal{M}(\lambda^*)\)). Namely, let \(f\) be the Cantor staircase (function), and let \(C\) be the Cantor set. Characterize the sets \(f\left(C^{\complement }\right)\) and \(f(C)\).
Consider the function \(g(x) = \frac{f(x) + x}{2}\). Prove that \(g\) satisfies the previous item, but is additionally strictly increasing, i.e., bijective on \([0,1]\).
Prove that the function \(h = g^{-1}\), where \(g\) is obtained in the previous item, is continuous and strictly increasing.
Use Vitaliβs construction to prove that there exists a non-Lebesgue-measurable \(B \subset g(C)\).
Let \(A := h(B) = g^{-1}(B)\), where \(B, h, g\) are as in the previous items. Prove that \(A\) is Lebesgue measurable.
Prove that \(A\) from the previous item is not Borel. Hint: a continuous strictly increasing function maps Borel sets to Borel sets.
Measurable functions and arithmetic operations
Let \(f: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) be a measurable function. Prove that \[
h(\omega ) = \begin{cases} \frac{1}{f(\omega )}, & f(\omega ) \neq 0 \\ c, & f(\omega ) = 0 \end{cases}
\] is measurable. Here \(c \in \mathbb {R}\) is an arbitrary constant.
Let \(f,g: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) be measurable functions. Prove that the event \(\left\{ f > g\right\}\) is measurable, i.e., that \[
\left\{ f > g\right\} = \left\{ \omega \; : \; f(\omega ) > g(\omega )\right\} \in \mathcal{F}
\]
The set {f>g} as a countable union β_{qββ}({f>q}β©{g<q}): move the slider qβ and find a rational qβ for which the point (f(Οβ),g(Οβ)) falls into the rectangle {f>qβ}β©{g<qβ} (it and the point turn green) β such a qβ must exist, since the point lies above the diagonal f>g.
Note that
\[
\left\{ f > g\right\} = \left\{ \omega \; : \; f(\omega ) > g(\omega )\right\} = \bigcup _{q \in \mathbb {Q}} \bigg[\left\{ \omega \; : \; f(\omega ) > q\right\} \cap \left\{ \omega \; : \; q > g(\omega )\right\} \bigg]
\]
Indeed, for an arbitrary \(\omega_0 \in \left\{ f > g\right\}\) there is a rational number \(q_0 = q(\omega_0)\) such that \(f(\omega_0) > q_0 > g(\omega_0)\) (since the rational numbers are dense in \(\overline{\mathbb {R}}\)). Hence
\[
\left\{ f > g\right\} \subset \bigcup _{q \in \mathbb {Q}} \bigg[\left\{ \omega \; : \; f(\omega ) > q\right\} \cap \left\{ \omega \; : \; q > g(\omega )\right\} \bigg]
\]
The reverse inclusion
\[
\left\{ f > g\right\} \supset \bigcup _{q \in \mathbb {Q}} \bigg[\left\{ \omega \; : \; f(\omega ) > q\right\} \cap \left\{ \omega \; : \; q > g(\omega )\right\} \bigg]
\]
is obvious.
We have
\[
\begin{align} \left\{ f > g\right\} & = \bigcup _{q \in \mathbb {Q}} \bigg[\left\{ \omega \; : \; f(\omega ) > q\right\} \cap \left\{ \omega \; : \; q > g(\omega )\right\} \bigg] = \\ & = \bigcup _{q \in \mathbb {Q}}\bigg[\underbrace{f^{-1}\left((q, +\infty ]\right)}_{\in \mathcal{F}} \cap \underbrace{g^{-1}\left([-\infty ,q)\right)}_{\in \mathcal{F}} \bigg] \in \mathcal{F} \end{align}
\]
We used the measurability of \(f\), \(g\), as well as the closedness of the sigma-algebra under countable intersections.
Let \(f,g: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) be measurable functions. Let \[
h(\omega ) = \begin{cases} f(\omega ) + g(\omega ), & f(\omega ) + g(\omega ) \text{ is defined} \\ c, & \text{ otherwise}\end{cases}
\] where \(c\in \overline{\mathbb {R}}\) is an arbitrary fixed number. Prove that \(h\) is measurable.
An indeterminacy can arise when \(f(\omega ) = +\infty\), \(g(\omega ) = -\infty\).
Let \(f,g: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) be measurable functions. Prove that \(\min (f,g), \max (f,g)\) are measurable functions.
The following problem shows that if \(f\) is measurable, \(g = f\) almost everywhere (i.e., the set of \(\omega\) where they are not equal has measure zero), and the space is complete, then \(g\) is also measurable.
Let \((\Omega , \mathcal{F}, \mu )\) be a measurable space with a complete measure \(\mu\). Let \(f: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) be a measurable function, and let \(g: \Omega \to \mathbb {R}\) be a function such that \(\left\{ f\neq g\right\} = \left\{ \omega \in \Omega : f(\omega ) \neq g(\omega )\right\} \in \mathcal{F}\) and \(\mu (\left\{ f \neq g\right\} ) = 0\). Prove that \(g\) is an \(\mathcal{F}|\mathscr {B}\left(\overline{\mathbb {R}}\right)\)-measurable function.
Show that the completeness condition on the measure cannot be dropped in the previous problem. That is, give an example of a measure space \((\Omega , \mathcal{F}, \mu )\) and functions \(f,g: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right))\) such that \(f=g\) a.e., while \(f\) is measurable but \(g\) is not.
Sequences of measurable functions
Theorem 1 Let \(f_n: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right)), n \in \mathbb {N}\) be a sequence of \(\mathcal{F}|\mathscr {B}\left(\overline{\mathbb {R}}\right)\)-measurable functions. Then
- The functions
\[
\sup _n f_n, \quad \inf _n f_n, \quad \varliminf _{n \to \infty } f_n, \quad \varlimsup _{n \to \infty } f_n
\]
are \(\mathcal{F}|\mathscr {B}\left(\overline{\mathbb {R}}\right)\)-measurable;
\(\left\{ \exists \lim_{n \to \infty }f_n\right\} = \left\{ \omega \in \Omega : \lim_{n\to \infty }f_n(\omega ) \text{ exists}\right\} \in \mathcal{F}\);
Let
\[
h(\omega ) := \begin{cases} \lim \limits _{n\to \infty }f_n(\omega ), & \lim \limits _{n\to \infty }f_n(\omega )\text{ exists} \\ c, & \text{ otherwise}\end{cases}
\]
where \(c \in \overline{\mathbb {R}}\) is an arbitrary fixed number. Then the function \(h\) is \(\mathcal{F}|\mathscr {B}\left(\overline{\mathbb {R}}\right)\)-measurable.
Prove item 2 of the theorem.
Use the Cauchy criterion.
We have
\[
\begin{align} \left\{ \exists \lim \limits _{n \to \infty } f_n\right\} & = \left\{ \omega \in \Omega : \lim _{n\to \infty }f_n(\omega ) \text{ exists}\right\} = \\ & = \left\{ \forall k \in \mathbb {N} \; \exists N \in \mathbb {N}\; :\; \forall m,n \geq N \hookrightarrow \left|f_n - f_m\right| < \frac{1}{k}\right\} = \\ & =\bigcap _{k=1}^{\infty }\left\{ \exists N\; :\; \forall m,n \geq N \hookrightarrow \left|f_n - f_m\right| < \frac{1}{k}\right\} = \\ & = \bigcap _{k=1}^{\infty } \bigcup _{N=1}^{\infty } \bigcap _{m,n = N}^{\infty } \left\{ \left|f_n - f_m\right| < \frac{1}{k}\right\} = \\ & = \bigcap _{k=1}^{\infty } \bigcup _{N=1}^{\infty } \bigcap _{m,n = N}^{\infty } \left(\left\{ f_{n} - f_{m} < \frac{1}{k}\right\} \cap \left\{ f_{n} - f_{m} > -\frac{1}{k}\right\} \right) \end{align}
\]
(a universal quantifier turns into an intersection, an existential quantifier turns into a union). Next, the sets
\[
\left\{ f_{n} - f_{m} < \frac{1}{k}\right\} , \left\{ f_{n} - f_{m} > -\frac{1}{k}\right\} \in \mathcal{F}
\]
i.e., they are measurable (see Example 2, where this is proved). \(\bigcap_{k=1}^{\infty } \bigcup_{N=1}^{\infty } \bigcap_{m,n = N}^{\infty }\) are countable intersections and unions. Hence,
\[
\left\{ \exists \lim \limits _{n \to \infty } f_n\right\} \in \mathcal{F}
\]
It is important that all unions and intersections range over at most countable collections of events.
Using the first 2 items, prove item 3 of the theorem.
Give an example of an uncountable collection \(f_e: (\Omega , \mathcal{F}) \to (\overline{\mathbb {R}}, \mathscr {B}\left(\overline{\mathbb {R}}\right)), e \in E\) of measurable functions such that the function \(\sup_{e \in E}f_e\) is not measurable.
βAlmost everywhereβ
Definition 3 Let \((\Omega , \mathcal{F}, \mu )\) be a measure space. We say that some property holds for almost all outcomes \(\omega\) (also said to hold \(\mu\)-almost everywhere, abbreviated \(\mu\)-a.e. or simply a.e.), if the measure of the outcomes where the property fails is zero. For example, one can say that almost all numbers in \(\mathbb {R}\) (taking Lebesgue measure on \(\mathbb {R}\) unless stated otherwise) are irrational, since the measure of the rational numbers is zero.
Define a function \(f(x)\) on the interval \([0, 1]\) (with Lebesgue measure) as follows. If \(x=\) \(\overline{0, n_{1} n_{2} n_{3} \cdots }\) is the decimal expansion of \(x\), let \(f(x)=\max_{i} n_{i}\). Prove that \(f(x)\) is measurable and is almost everywhere constant.
Let \(f : \mathbb {R} \to \overline{\mathbb {R}}\). On the domain we take Lebesgue measure. Consider 2 conditions:
\(f\) is continuous almost everywhere;
there exists a continuous function \(g: \mathbb {R} \to \overline{\mathbb {R}}\) such that \(f=g\) almost everywhere. Give examples showing that neither condition follows from the other.
2 new modes of convergence for functions: almost everywhere and in measure
In this section, instead of the extended real line \(\overline{\mathbb {R}}\) the functions will take values in \(\mathbb {R}\), in order to avoid indeterminate forms such as \(+\infty - (+\infty )\).
Definition 4 Let \((\Omega , \mathcal{F}, \mu )\) be a measure space. Let \(f, f_1, f_2, \dots : (\Omega , \mathcal{F}) \to (\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))\) be a collection of \(\mathcal{F}|\mathscr {B}\left(\mathbb {R}\right)\)-measurable functions. We say that \(f_n\) converges to \(f\) almost everywhere with respect to the measure \(\mu\), or \(\mu\)-a.e. (or simply a.e., if it is clear which measure is meant), if \[
\mu \left(\left\{ \lim f_n \neq f\right\} \right) = \mu \left(\left\{ \omega \in \Omega : \lim _{n \to \infty } f_n(\omega ) \neq f(\omega )\right\} \right) = 0
\] The notation β\(\lim_{n \to \infty } f_n(\omega ) \neq f(\omega )\)β means that either the limit does not exist, or it exists but is not equal to \(f(\omega )\).
For this definition to be well posed we need the measurability of \(\left\{ \lim f_n \neq f\right\}\), i.e., we need \(\left\{ \omega \in \Omega : \lim_{n \to \infty } f_n(\omega ) \neq f(\omega )\right\} \in \mathcal{F}\). Note that
\[
\begin{align} \left\{ \omega \in \Omega : \lim _{n \to \infty } f_n(\omega ) = f(\omega )\right\} & = \left\{ \omega \in \Omega : \; \forall m \in \mathbb {N} \, \exists N \in \mathbb {N} \, \forall n \geq N \, \hookrightarrow \, \left|f(\omega ) - f_{n}(\omega )\right| \leq \frac{1}{m}\right\} = \\ & = \bigcap _{m=1}^\infty \bigcup _{N=1}^\infty \bigcap _{n=N}^\infty \underbrace{\left\{ \omega : \left|f(\omega ) - f_n(\omega )\right| \leq \frac{1}{m}\right\} }_{\in \mathcal{F}} \in \mathcal{F} \end{align}
\]
Hence,
\[
\left\{ \lim f_n \neq f\right\} = \left\{ \lim f_n = f\right\} ^{\complement } \in \mathcal{F}
\]
Definition 5 Let \((\Omega , \mathcal{F}, \mu )\) be a measure space. Let \(f, f_1, f_2, \dots : (\Omega , \mathcal{F}) \to (\mathbb {R}, \mathscr {B}\left(\mathbb {R}\right))\) be a collection of \(\mathcal{F}|\mathscr {B}\left(\mathbb {R}\right)\)-measurable functions. We say that \(f_n\) converges to \(f\) in measure \(\mu\) if \[
\mu \left(\omega : \left|f_n(\omega ) - f(\omega )\right| > \varepsilon \right) \xrightarrow [n \to \infty ]{} 0, \quad \forall \varepsilon > 0
\]
In probability theory, as you may recall, convergence in measure was called convergence in probability. Convergence in probability is a special case of convergence in measure, in which the measure is a probability measure.
Let \[
f_n(x) = \frac{nx}{1 + n^2x^2}, \qquad g_n(x) = \frac{n^2x}{n^4 + x^2}
\] Do the function sequences below converge with respect to Lebesgue measure, or almost everywhere, on the interval \(I\)?
\(f_n, \quad I = (0,1)\)
\(f_n, \quad I = (1,+\infty )\)
\(g_n, \quad I = (0,1)\)
\(g_n, \quad I = (1,+\infty )\)
Let \(f_n \to f\) almost everywhere, \(f_n \to g\) almost everywhere. Prove that then \(f=g\) almost everywhere.
Let \(f_n \to f\) in measure, \(f_n \to g\) in measure. Prove that then \(f=g\) almost everywhere.
Construct a sequence \((f_n)\) of functions on \([0,1]\) such that \(f_n \to 0\) in (Lebesgue) measure, but \(\varlimsup_{n \to \infty } f_n = +\infty\) almost everywhere and \(\varliminf_{n \to \infty } f_n = -\infty\) almost everywhere.
Let a sequence of nonnegative functions \(\left\{ f_{n}(x)\right\}_{n=1}^{\infty }\) converge in measure to \(f(x)\) on \(A\). Prove that \(f(x) \geqslant 0\) a.e. on \(A\).
Theorem 2 (F. Riesz) If \((f_n)\) converges to \(f\) in measure, then there is a subsequence \((f_{n_k})_{k=1}^\infty\) that converges to \(f\) almost everywhere.
Theorem 3 (H. Lebesgue) If \(\mu \left(\Omega \right) < +\infty\), then convergence a.e. implies convergence in measure.
Show that, in general, convergence a.e. does not imply convergence in measure when the measure is \(\sigma\)-finite.
Egorovβs theorem
A pointwise convergent sequence of functions need not converge uniformly. However, the following result asserts that a pointwise convergent sequence of functions on a measure space of finite total measure converges almost uniformly, in the sense that it converges uniformly except on a set that can be made to have arbitrarily small measure.
Theorem 4 (D. Egorov) Let \(\mu \left(\Omega \right) < +\infty\) and let \(f_n\) converge to \(f\) almost everywhere. Then \(\forall \varepsilon > 0\) there exists \(B_\varepsilon \in \mathcal{F}\) such that \(f_n \to f\) uniformly on \(B_\varepsilon\) and \(\mu \left(B_\varepsilon^\complement \right) < \varepsilon\).
Show that the condition \(\mu \left(\Omega \right) < +\infty\) cannot be dropped in Egorovβs theorem. Show, for example, that the conclusion of Egorovβs theorem fails for Lebesgue measure on \(\mathbb {R}\).
Lebesgue measurable functions
The passage from Borel measurable functions to Lebesgue measurable functions is the work of the devil. Donβt even think about it!
β Barry Simon (winner of the American Mathematical Societyβs Steele Prize) in his five-volume βA Comprehensive Course in Analysisβ
Although there exist Lebesgue measurable sets that are not Borel, you are unlikely to encounter them. Likewise, a Lebesgue measurable function that is not Borel measurable is unlikely to arise in your work. An excellent way to avoid potential confusion is to consider only Borel measurable functions.
The following result asserts that if we adhere to the philosophy that what happens on a set of outer measure \(0\) does not matter much, then we may restrict our attention to Borel measurable functions.
Theorem 5 (every Lebesgue measurable function is almost Borel measurable.) Suppose \(f: \mathbb {R} \rightarrow \mathbb {R}\) is a Lebesgue measurable function. Then there exists a Borel measurable function \(g: \mathbb {R} \rightarrow \mathbb {R}\) such that \[
\lambda \left(g \neq f\right) = \lambda \left(\left\{ x \in \mathbb {R} \; : \; g(x) \neq f(x)\right\} \right) = 0
\]