Chapter 3: Lebesgue Delta Measurable Functions

Proof

Note that for any Borel set \(E\subset \mathcal{B}\) we have the following cases 1. \(0, 1\in E\), 2. \(0, 1\notin E\), 3. \(0\in E, 1\notin E\), 4. \(0\notin E, 1\in E\). Therefore \(f^{-1}(E)\) is either \(\mathbb{T}\), \(\emptyset\), \(\mathbb{T}\backslash A\), or \(A\). Thus, the assertion follows.

Proof

For any Borel measurable set \(B\subset \mathcal{B}\), we have \(g^{-1}(B)\subset \mathcal{B}\). Hence,

\[ \begin{aligned} (g\circ f)^{-1}(B)=& f^{-1}(g^{-1}(B))\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag \]

This completes the proof.

Proof

We prove only the equivalence of some of the statements above which can then be combined together to give the full equivalence. \(\mathbf{(1)\Longrightarrow (2)}\) By the definition for Lebesgue delta measurable functions, we get that \(f^{-1}(I)\subset \mathbb{M}(m^*)\) for any open interval \(I\subset \mathbb{R}\). \(\mathbf{(2)\Longrightarrow (1)}\) By the fact that the Borel \(\sigma\)-algebra is generated by the open sets of \(\mathbb{R}\) and the preimage operation preserves countable unions, intersections, and complements, we get that \(f^{-1}(A)\subset \mathbb{M}(m^*)\) for any \(A\subset \mathcal{B}\). Now, applying the definition for Lebesgue delta measurable functions, we conclude that the function \(f\) is Lebesgue delta measurable. \(\mathbf{(2)\Longrightarrow (3)}\) For a fixed \(a\in \mathbb{R}\), define the open intervals

\[ I_n=\left(-\infty, a+\frac{1}{n}\right].\notag\]

Then, the set \((-\infty, a]\). can be written as the countable intersection

\[ (-\infty, a]=\bigcap\limits_{n=1}^\infty I_n.\notag\]

Hence, we have

\[\begin{aligned} f^{-1}\left((-\infty, a]\right)=& f^{-1}\left(\bigcap\limits_{n=1}^\infty I_n\right)\\ =& \bigcap\limits_{n=1}^\infty f^{-1}(I_n)\\ \subset& \mathbb{M}(m^*). \end{aligned} \notag\]

\(\mathbf{(3)\Longrightarrow (2)}\) We first show that we can produce any open interval \((a, b)\) for \(a, b\in \mathbb{R}\) with \(a<b\) from these half-infinity intervals by countable unions, intersection, and complements. For finite \(a, b\in \mathbb{R}\), we have that

\[ \begin{aligned} (a, b]=& (-\infty, b]\backslash (-\infty, a],\\ \{b\}=& \bigcap\limits_{n=1}^\infty. \left(b-\frac{1}{b}, b\right]\\ =& \bigcap \left((-\infty, b]\backslash \left(-\infty, b-\frac{1}{n}\right]\right). \end{aligned} \notag\]

Therefore, the set \((a, b)=(a, b)\backslash \{b\}\) can be made up of countable unions, intersections, and complements of intervals of the form \((-\infty, a)\). Next, the intervals \((-\infty, a)\) and \((a, \infty)\) can be represented as follows

\[(-\infty, a)=\left(\ldots \cup (a-3, a-2]\cup (a-2, a-1]\right)\cup (a-1, a)\notag\]

and

\[ (a, \infty)=(a, a+1)\cup \left([a+1, a+2)\cup [a+2, a+3)\cup \ldots\right). \notag\]

Thus, by using the fact that the inverse functions preserve unions, intersections, and complements, we conclude the result. \(\mathbf{(2)\Longrightarrow (4)}\) This is trivially true since \((-\infty, a)\) are open intervals in \(\mathbb{R}\). \(\mathbf{(4)\Longrightarrow (2)}\) We need to show that for any \(a, b\in \mathbb{R}\) with \(a<b\), the interval \((a, b)\) can be constructed by countable intersections. For finite \(a, b\in \mathbb{R}\) with \(a<b\), we note that

\[ \begin{aligned} {[a, b)}=(-\infty, b)\backslash (-\infty, a) \end{aligned} \notag\]

and

\[ \begin{aligned} \{a\}=& \bigcap\limits_{n=1}^\infty \left [a, a+\frac{1}{n}\right)\\ \\ =& \bigcap\limits_{n=1}^\infty \left(\left(-\infty, a+\frac{1}{n}\right)\backslash (-\infty, a)\right). \end{aligned} \notag\]

Therefore, the set

\[ (a, b)=(-\infty, b)\backslash (-\infty, a)\cap \{a\}^c \notag\]

is made up of countable unions, intersections, and complements of intervals of the forms \((-\infty, a)\) and \((a, \infty)\). Thus, by using the fact that the inverse functions preserve intersections, we conclude the result. \(\mathbf{(3)\Longleftrightarrow (6)}\)

Note that for any \(a\in \mathbb{R}\), we have \((a, \infty)={(-\infty, a]}^c.\) Thus,

\[ \begin{aligned} f^{-1}((a, \infty))=& f^{-1}\left({(-\infty, a]}^c\right)\\ =& \mathbb{T}\backslash f^{-1}({(-\infty, a]}). \end{aligned} \notag\]

This implies that \(f^{-1}((-\infty, a))\in \mathbb{M}(m^*)\) for any \(a\in \mathbb{R}\) if and only if \(f^{-1}({(-\infty, a]})\in \mathbb{M}(m^*)\) for any \(a\in \mathbb{R}\). \(\mathbf{(4)\Longleftrightarrow (5)}\) Note that for any \(a\in \mathbb{R}\), we have

\[ [a, \infty)=(-\infty, a)^c. \notag\]

Hence,

\[ \begin{aligned} f^{-1}({[a, \infty)})=& f^{-1}\left({(-\infty, a)}^c\right)\\ \\ =& \mathbb{T}\backslash f^{-1}((-\infty, a)). \end{aligned} \notag\]

From here, we get that \(f^{-1}((-\infty, a))\in \mathbb{M}(m^*)\) for any \(a\in \mathbb{R}\) if and only if \(f^{-1}([a, \infty))\). This completes the proof. \end{proof}

Proof

1. If \(\lambda=0\), then there is nothing to check. Suppose that \(\lambda>0\). Since \(f: \mathbb{T}\to \mathbb{R}\) is Lebesgue delta measurable, then all preimages of the intervals \((-\infty, b)\) for any \(b\in \mathbb{R}\) are contained in \(\mathbb{M}(m^*)\). In other words, we know that the sets

\[ \{ t\in \mathbb{T}: f(t)<b\} \notag \]

lie in \(\mathbb{M}(m^*)\) for any \(b\in \mathbb{R}\). To show that \(\lambda f: \mathbb{T}\to \mathbb{R}\) is Lebesgue delta measurable, we need to show that

\[ \{t\in \mathbb{T}: \lambda f(t)<c\}\subset \mathbb{M}(m^*) \notag\]

for any \(c\in \mathbb{R}\). However, for each \(c\in \mathbb{R}\), this is simply

\[ \begin{aligned} \{t\in \mathbb{T}: \lambda f(t)<c\}=& \left\{t\in \mathbb{T}: f(t)<\frac{c}{\lambda}\right\}\\ =& \{t\in \mathbb{T}: f(t)<b\}\quad \mbox{for}\quad b=\frac{c}{\lambda}\in \mathbb{R}, \end{aligned} \notag\]

which we know lies in \(\mathbb{M}(m^*)\). Finally, for the case \(\lambda<0\), we have

\[ \left\{t\in \mathbb{T}: \lambda f(t)<c\right\}=\left\{t\in \mathbb{T}: f(t)>\frac{c}{\lambda}\right\}, \notag\]

which is also contained in \(\mathbb{M}(m^*)\). 2. We need to check that for all \(c\in. \mathbb{R}\), the sets

\[ \{t\in \mathbb{T}: f(t)+g(t)>c\} \notag\]

lie in \(\mathbb{M}(m^*)\). Observe that

\[ \{t\in \mathbb{T}: f(t)+g(t)>c\}=\bigcup\limits_{q\in \mathbb{Q}}\left(\{t\in \mathbb{T}: f(t)>q\}\cap \{t\in \mathbb{T}: g(t)>c-q\}\right). \notag\]

Since \(f, g: \mathbb{T}\to \mathbb{R}\) are Lebesgue delta measurable we know that the sets

\[ \{t\in \mathbb{T}: f(t)>q\}\quad \mbox{and}\quad \{t\in \mathbb{T}: g(t)>c-q\} \notag\]

lie in \(\mathbb{M}(m^*)\) for each \(c\in \mathbb{R}\) and \(q\in \mathbb{Q}\). Since \(\mathbb{M}(m^*)\) is a \(\sigma\)-algebra, this countable union is also in \(\mathbb{M}(m^*)\). Similarly, we have

\[ \begin{aligned} \{t\in \mathbb{T}: f(t)-g(t)>c\}=&\bigcup\limits_{q\in \mathbb{Q}}\left(\{t\in \mathbb{T}: f(t)>q\}\cap \{t\in \mathbb{T}: g(t)<q-c\}\right)\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

So, the function \(f-g\) is also Lebesgue delta measurable. 3. To show that $fg: \mathbb{T}\to\mathbb{R}\) is Lebesgue delta measurable, firstly we will show that the function \(f^2: \mathbb{T}\to \mathbb{R}\) is Lebesgue delta measurable. This is equivalent to showing that the set

\[ \{t\in \mathbb{T}: f(t)^2<c\} \notag\]

is in \(\mathbb{M}(m^*)\) for any \(c\in \mathbb{R}\). We split this into two cases. (a) For \(c<0\), we get

\[ \begin{aligned} \{t\in \mathbb{T}: f(t)^2<c\}=&\emptyset\\ \\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

(b) For \(c\geq 0\), we find

\[ \{t\in \mathbb{T}: f(t)^2<c\}=\{t\in \mathbb{T}: (f(t)-\sqrt{c})(f(t)+\sqrt{c})<0\} \notag\]

\[ \begin{aligned} =& \left( \{t\in \mathbb{T}: f(t)>-\sqrt{c}\}\cap \{t\in \mathbb{T}: f(t)<\sqrt{c}\}\right)\\ &\bigcup\left(\{t\in \mathbb{T}: f(t)<-\sqrt{c}\}\cap \{t\in \mathbb{T}: f(t)>\sqrt{c}\}\right), \end{aligned} \notag\]

which lies in \(\mathbb{M}(m^*)\) since \(f\) is Lebesgue delta measurable.

To show that \(fg:\mathbb{T}\to\mathbb{R}\) is Lebesgue delta measurable, we note that

\[ fg=\frac{1}{4}\left((f+g)^2-(f-g)^2\right). \notag\]

Since \(f\) and \(g\) are Lebesgue delta measurable, necessarily \(f\pm g\) are Lebesgue delta measurable by the second assertion. Hence, \(\frac{1}{4}(f\pm g)^2\) are Lebesgue delta measurable by the above and the first assertion. Finally, we conclude that their difference, which is \(fg\), is also Lebesgue delta measurable by the second assertion.

4. To show that \(\frac{f}{g}:\left\{t\in \mathbb{T}: g(t)\ne 0\right\}\) is Lebesgue delta measurable, it suffices to show that the function \(\frac{1}{g}:\{t\in \mathbb{T}: g(t)\ne 0\}\to \mathbb{R}\) is Lebesgue delta measurable. We check for different cases of the value \(c\).

(a) For \(c>0\), we have

\[ \begin{aligned} \left\{t\in \mathbb{T}: \frac{1}{g(t)}<c\right\}=& \left\{t\in \mathbb{T}: \frac{1}{c}<g(t)\right\}\bigcap \{t\in \mathbb{T}: g(t)<0\}\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

(b) For \(c=0\), we have

\[ \begin{aligned} \left\{t\in \mathbb{T}: \frac{1}{g(t)}<0\right\}=& \{t\in \mathbb{T}: g(t)<0\}\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

(c) For \(c>0\), we get

\[ \begin{aligned} \left\{t\in \mathbb{T}: \frac{1}{g(t)}<c\right\}=& \left\{t\in \mathbb{T}: \frac{1}{c}<g(t)\right\}\bigcup \{t\in \mathbb{T}: g(t)<0\}\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

Thus, the function \(\frac{1}{g}\) is also Lebesgue delta measurable. Since the product of Lebesgue delta measurable functions is also Lebesgue delta measurable, by the third assertion, we conclude that \(\frac{f}{g}\) is also Lebesgue delta measurable.

5. Note that

\[ \{t\in \mathbb{T}: \max\{f, g\}(t)<c\}=\{t\in \mathbb{T}: f(t)<c\}\cap \{t\in \mathbb{T}: g(t)<c\} \notag\]

and

\[ \{t\in \mathbb{T}: \min\{f, g\}(t)>c\}= \{t\in \mathbb{T}: f(t)>c\}\cap \{t\in \mathbb{T}: g(t)>c\} \notag\]

for any \(c\in \mathbb{R}\). Thus, the functions \(\max\{f, g\}\) and \(\min\{f, g\}\) are Lebesgue delta measurable.

6. Since \(0\) is Lebesgue delta measurable, by the previous assertion, we conclude that \(f^+\) and \(f^-\). are Lebesgue delta measurable. Next,

\[ \{t\in \mathbb{T}: |f(t)|<c\}=\{t\in \mathbb{T}: f(t)<c\}\cap \{t\in \mathbb{T}: f(t)>-c\}, \notag\]

whereupon we conclude that \(|f|\) is Lebesgue delta measurable. This completes the proof.

Proof

By assumption, we have that \(f: \mathbb{T}\to \overline{\mathbb{R}}\) is Lebesgue delta measurable. Then, we have \(f^{-1}(\{-\infty\})\subset \mathbb{M}(m^*)\). Since \((-\infty, a]\subset \mathcal{B}\), we get that \(f^{-1}((-\infty, a])\subset \mathbb{M}(m^*)\). Hence,

\[ \begin{aligned} f^{-1}([-\infty, a])=& f^{-1}(\{-\infty\}\cup (-\infty, a])\\ =& f^{-1}(\{-\infty\})\cup f^{-1}((-\infty, a])\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

\(\mathbf{(2)\Longrightarrow (1)}\)

By assumption, we have \(f^{-1}([-\infty, a])\subset \mathbb{M}(m^*)\). We first show that \(f^{-1}(\{-\infty\}), f^{-1}(\{\infty\})\subset \mathbb{M}(m^*)\).

(a) Note that

\[ \{-\infty\}=\bigcap\limits_{n\in \mathbb{Z}}I_n. \notag\]

Then

\[ \begin{aligned} f^{-1}( \{-\infty\})=& f^{-1}\left(\bigcap\limits_{n\in \mathbb{Z}}I_n\right)\\ =& \bigcap\limits_{n\in \mathbb{Z}}f^{-1}(I_n)\\ \subset & \mathbb{M}(m^*). \end{aligned} \notag\]

Now, using De Morgans law, we have

\[ \begin{aligned} \{\infty\}=& \overline{\mathbb{R}}\backslash \bigcup\limits_{n\in \mathbb{Z}}I_n\\ =& \bigcap\limits_{n\in \mathbb{Z}}I_n^c. \end{aligned} \notag\]

Hence,

\[ \begin{aligned} f^{-1}(\{\infty\})=& f^{-1}\left(\bigcap\limits_{n\in \mathbb{Z}}I_n^c\right)\\ =& \bigcap\limits_{n\in \mathbb{Z}}f^{-1}(I_n^c)\\ =& \bigcap\limits_{n\in \mathbb{Z}}\left(\mathbb{T}\backslash f^{-1}(I_n)\right)\\ =& \mathbb{T} \backslash \bigcup\limits_{n\in \mathbb{Z}}f^{-1}(I_n)\\ \subset& \mathbb{M}(m^*). \end{aligned} \notag\]

Note that

\[ (-\infty, a]=[-\infty, a]\backslash \{-\infty\}. \notag\]

Then

\[ \begin{aligned} f^{-1}((-\infty, a])=& f^{-1}([-\infty, a]\backslash \{-\infty\})\\ =& f^{-1}([-\infty, a])\backslash f^{-1}(\{-\infty\})\\ \subset& \mathbb{M}(m^*). \end{aligned} \notag\]

From here, we get that for any \(A\subset \mathcal{B}\) one has \(f^{-1}(A)\subset \mathbb{M}(m^*)\). Therefore \(f\) is Lebesgue delta measurable. This completes the proof. \end{proof}

Proof

\[ \begin{aligned} \{t\in \mathbb{T}: \sup\limits_{n\in \mathbb{N}}f_n(t)\leq c\}=& \bigcap\limits_{n\in \mathbb{N}}\{t\in \mathbb{T}: f_n(t)\leq c\}\\ \subset & \mathbb{M}(m^*)\quad \mbox{for any}\quad c\in \mathbb{R} \end{aligned} \notag\]

and

\[ \begin{aligned} \{t\in \mathbb{T}: \inf\limits_{n\in \mathbb{N}}f_n(t)\leq c\}=& \bigcup\limits_{n\in \mathbb{N}}\{t\in \mathbb{T}: f_n(t)\leq c\}\\ \subset & \mathbb{M}(m^*)\quad \mbox{for any}\quad c\in \mathbb{R}. \end{aligned} \notag\]

Therefore \(\sup\limits_{n\in \mathbb{N}}f_n\) and \(\inf\limits_{n\in \mathbb{N}}f_n\) are Lebesgue delta measurable.

2. Define the sequences \(\{\sup\limits_{m\geq n}f_m(t)\}_{n\in \mathbb{N}}\) and \(\{\inf\limits_{m\geq n}f_n(t)\}_{n\in \mathbb{N}}\) for \(t\in \mathbb{T}\). By the first assertion, we have that the functions \(\sup\limits_{m\geq n}f_m(t)\) and \(\inf\limits_{m\geq n}f_n(t)\), \(n\in \mathbb{N}\), are Lebesgue delta measurable for any \(n\in \mathbb{N}\). Furthermore, since

\[\begin{aligned} \limsup\limits_{n\to\infty}f_n(t)=&\inf\limits_{n\in \mathbb{N}}\left(\sup\limits_{m\geq n}f_m(t)\right),\\ \liminf\limits_{n\to \infty}f_n(t)=& \sup\limits_{n\in \mathbb{N}}\left(\inf\limits_{m\geq n}f_n(t)\right),\quad t\in \mathbb{T},\end{aligned}\notag\]

applying the first assertion, we get that \(\limsup\limits_{n\to \infty} f_n\) and \(\liminf\limits_{n\to \infty} f_n\) are Lebesgue delta measurable. This completes the proof. \end{proof}