The region under the graph
Definition (Region under the graph and graph of a function)
Let \(f : E \to \overline{\mathbb{R}}\) be measurable with \(E \subseteq \mathbb{R}^{d}\) measurable and \(f \geq 0\). We define the region under the graph of \(f\) over \(E\) as
\[R(f,E) = \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in E,\ 0 \leq y \leq f(x) \},\]and the graph of \(f\) over \(E\) as
\[\Gamma(f,E) = \{ (x, f(x)) :\ x \in E \}.\]The question arises: is \(R(f,E)\) measurable?
Example (The region under a multiple of an indicator)
Let us analyse the case
\[f(x) = a \mathbf{1}_{A}(x)\]with \(A \subseteq E\) measurable and \(a > 0\). Then we have that
\[\begin{aligned} R(f,E) &= \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in E,\ 0 \leq y \leq f(x) \} \\ &= \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in A,\ 0 \leq y \leq a \} \cup \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in E \setminus A,\ y = 0 \} \\ &= \big( A \times [0,a] \big) \cup \big( (E \setminus A) \times \{0\} \big). \end{aligned}\]That is, the measurability of \(R(f,E)\) in this case reduces to the measurability of products of the form \(A \times [0,a]\).
Lemma (Measure of the product of a measurable set with an interval)
Let \(A \subseteq \mathbb{R}^{d}\) be measurable and \(a \geq 0\). Then \(A \times [0,a] \subseteq \mathbb{R}^{d+1}\) is measurable and its measure is
\[m(A \times [0,a]) = a\, m(A),\]with the convention \(0 \cdot \infty = 0\).
Proof: The case of a box. If \(A = [a_{1},b_{1}] \times \dots \times [a_{d},b_{d}]\), then \(A \times [0,a]\) is a box in \(\mathbb{R}^{d+1}\) and the result follows from the formula for the volume of boxes. Checking the details is left as an exercise for the reader.
The case of an open set with \(m(A) < \infty\). If \(A\) is open, then there exist boxes \(I_{k}\) in \(d\) dimensions such that
\[A = \bigcup_{k=1}^{\infty} I_{k} \quad \text{with } I_{j}^{\circ} \cap I_{k}^{\circ} = \emptyset \text{ if } j \neq k.\]Then
\[A \times [0,a] = \bigcup_{k=1}^{\infty} I_{k} \times [0,a]\]is a measurable set, being a countable union of boxes. Since
\[(I_{k} \times [0,a])^{\circ} \cap (I_{j} \times [0,a])^{\circ} = \emptyset, \quad k \neq j,\]and the measure of a countable union of boxes with pairwise disjoint interiors is the sum of their measures, we have that
\[m(A \times [0,a]) = \sum_{i=1}^{\infty} m(I_{i} \times [0,a]) = a \sum_{i=1}^{\infty} m(I_{i}) = a\, m(A).\]The case of a \(G_{\delta}\) with \(m(A) < \infty\). Now let
\[A = \bigcap_{j=1}^{\infty} G_{j}\]be a \(G_{\delta}\) with each \(G_{i}\) open and \(G_{i+1} \subseteq G_{i}\). We may assume monotonicity because, replacing \(G_{j}\) by \(G_{1} \cap \dots \cap G_{j}\), which is still open, the total intersection does not change. Moreover, since \(m(A) < \infty\), by outer regularity there exists an open set \(G \supseteq A\) with \(m(G) < \infty\); intersecting each \(G_{j}\) with \(G\) we may also suppose that \(m(G_{1}) < \infty\). Hence
\[A \times [0,a] = \bigcap_{j=1}^{\infty} G_{j} \times [0,a],\]with \(G_{i+1} \times [0,a] \subseteq G_{i} \times [0,a]\) measurable and of finite measure, since \(m(G_{i} \times [0,a]) = a\, m(G_{i}) \leq a\, m(G_{1}) < \infty\) by the previous case. Then, by continuity from above of the measure,
\[m(A \times [0,a]) = \lim_{i \to \infty} m(G_{i} \times [0,a]) = \lim_{i \to \infty} a\, m(G_{i}) = a\, m(A).\]The general measurable case with \(m(A) < \infty\). In the case where \(A\) is measurable with finite measure, we may write
\[A = H \setminus Z\]with \(H\) a \(G_{\delta}\) set such that \(A \subseteq H\), \(m(H) = m(A)\), and \(Z = H \setminus A\) of measure zero. Then, by the previous step, \(H \times [0,a]\) is measurable and
\[m(H \times [0,a]) = a\, m(H).\]It is left as an exercise to prove that
\[m_{e}(Z \times [0,a]) = 0.\]Then \(Z \times [0,a]\) is measurable with measure zero,
\[A \times [0,a] = (H \times [0,a]) \setminus (Z \times [0,a])\]is measurable, and
\[m(A \times [0,a]) = m(H \times [0,a]) = a\, m(H) = a\, m(A).\]The case \(m(A) = \infty\). Finally, if \(m(A) = \infty\), set
\[A_{k} = A \cap B(0,k).\]In this way
\[A = \bigcup_{k=1}^{\infty} A_{k},\]with \(A_{k} \subseteq A_{k+1}\) measurable and bounded, in particular of finite measure. By the previous arguments,
\[A_{k} \times [0,a] \subseteq A_{k+1} \times [0,a]\]are measurable sets such that
\[m(A_{k} \times [0,a]) = a\, m(A_{k}).\]Then \(A \times [0,a] = \bigcup_{k=1}^{\infty} A_{k} \times [0,a]\) is measurable and, by continuity from below,
\[m(A \times [0,a]) = \lim_{k \to \infty} m(A_{k} \times [0,a]) = \lim_{k \to \infty} a\, m(A_{k}) = a\, m(A).\]Exercise (The product of a null set with an interval is null)
Let \(Z \subseteq \mathbb{R}^{d}\) with \(m(Z) = 0\) and \(a \geq 0\). Show that \(m_{e}(Z \times [0,a]) = 0\).
The graph of a measurable function
Lemma (The graph of a measurable function has measure zero)
Let \(f : E \to \mathbb{R}\) be measurable, with \(E\) measurable. Then \(m_{e}(\Gamma(f,E)) = 0\).
Proof: Let us first assume that \(E\) has finite measure. Let \(\varepsilon > 0\) and, for each \(k \in \mathbb{Z}\),
\[E_{k} = \{ x \in E :\ k \varepsilon \leq f(x) < (k+1) \varepsilon \}.\]The \(E_{k}\) are measurable, pairwise disjoint, and
\[E = \bigcup_{k \in \mathbb{Z}} E_{k}.\]Note that
\[\Gamma(f, E_{k}) = \{ (x, f(x)) :\ x \in E_{k} \} \subseteq E_{k} \times [k\varepsilon, (k+1)\varepsilon).\]By the previous lemma and the invariance of outer measure under translations, we have that
\[m_{e}(\Gamma(f,E_{k})) \leq m_{e}\big( E_{k} \times [k\varepsilon,(k+1)\varepsilon) \big) \leq \varepsilon\, m(E_{k}).\]Therefore, by countable subadditivity of outer measure, it holds that
\[m_{e}(\Gamma(f,E)) = m_{e}\left( \bigcup_{k \in \mathbb{Z}} \Gamma(f,E_{k}) \right) \leq \sum_{k \in \mathbb{Z}} m_{e}(\Gamma(f,E_{k})) \leq \sum_{k \in \mathbb{Z}} \varepsilon\, m(E_{k}) = \varepsilon\, m(E),\]where the last equality uses that the \(E_{k}\) are pairwise disjoint. Since \(\varepsilon > 0\) is arbitrary and \(m(E) < \infty\), we conclude that \(m_{e}(\Gamma(f,E)) = 0\).
Finishing the proof in the case \(m(E) = \infty\) is left as an exercise (see the following exercise).
Exercise (The infinite measure case of the graph lemma)
Finish the proof of the previous lemma: show that if \(m(E) = \infty\), then also \(m_{e}(\Gamma(f,E)) = 0\).
Lemma (The region under a non-negative simple function is measurable)
Let \(\phi : E \to \mathbb{R}\) be simple, measurable and non-negative, say
\[\phi(x) = \sum_{i=1}^{m} a_{i} \mathbf{1}_{A_{i}},\]with \(a_{i} \neq a_{j}\) and \(A_{i} \cap A_{j} = \emptyset\) for \(i \neq j\), and \(E = \bigcup_{i=1}^{m} A_{i}\). Then \(R(\phi,E)\) is measurable.
Proof: We have that
\[R(\phi,E) = \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in E,\ 0 \leq y \leq \phi(x) \} = \bigcup_{j=1}^{m} \{ (x,y) \in \mathbb{R}^{d+1} :\ x \in A_{j},\ 0 \leq y \leq a_{j} \},\]that is, \(R(\phi,E) = \bigcup_{j=1}^{m} A_{j} \times [0,a_{j}]\), which is a finite union of measurable sets by the product lemma, and is therefore a measurable set.
Theorem (The region under the graph of a non-negative measurable function is measurable)
Let \(f : E \to \mathbb{R}\) be measurable with \(f \geq 0\) and \(E\) measurable. Then \(R(f,E) \subseteq \mathbb{R}^{d+1}\) is a measurable set.
Proof: Since \(f \geq 0\) is measurable, there exists an increasing sequence of simple, measurable and non-negative functions \(\{\phi_{k}\}_{k=1}^{\infty}\) satisfying
\[\lim_{k \to \infty} \phi_{k}(x) = f(x)\]on \(E\). Hence, if \(0 \leq y < f(x)\), since \(\phi_{k}(x) \to f(x) > y\), there exists \(\phi_{k}\) such that
\[0 \leq y \leq \phi_{k}(x).\]Then, separating the points with \(y < f(x)\) from those with \(y = f(x)\),
\[R(f,E) = \{ (x,y) :\ x \in E,\ 0 \leq y < f(x) \} \cup \{ (x,y) :\ x \in E,\ f(x) = y \},\]and by the above we obtain
\[R(f,E) = \bigcup_{k=1}^{\infty} R(\phi_{k},E) \cup \{ (x,y) :\ x \in E,\ f(x) = y,\ y \geq 0 \},\]since each \(R(\phi_{k},E) \subseteq R(f,E)\) because \(\phi_{k} \leq f\). Now, each \(R(\phi_{k},E)\) is measurable by the previous lemma, and the second set is contained in \(\Gamma(f,E)\), which has outer measure zero and is therefore measurable. We conclude that \(R(f,E)\) is measurable, being a countable union of measurable sets.
The definition of the integral
Definition (Lebesgue integral of a non-negative measurable function)
Given \(f : E \to \mathbb{R}\) measurable such that \(f \geq 0\), we define its Lebesgue integral as
\[\int_{E} f \, dx = m(R(f,E)).\]Proposition (The integral of a non-negative simple function)
Let \(\phi = \sum_{i=1}^{m} a_{i} \mathbf{1}_{A_{i}}\) be simple, measurable and non-negative, with the \(A_{i} \subseteq E\) measurable and pairwise disjoint. Then
\[\int_{E} \phi(x) \, dx = \sum_{i=1}^{m} a_{i}\, m(A_{i}).\]Proof: We may assume that \(a_{i} \neq 0\) for \(1 \leq i \leq m\), since the terms with \(a_{i} = 0\) contribute to neither side. Then
\[\begin{aligned} \int_{E} \phi(x)\, dx &= m\big( R(\phi,E) \big) \\ &= m\big( \{ \phi = 0 \} \times \{0\} \big) + m\left( \bigcup_{i=1}^{m} \{ (x,y) :\ x \in A_{i},\ 0 \leq y \leq a_{i} \} \right) \\ &= 0 + \sum_{i=1}^{m} a_{i}\, m(A_{i}), \end{aligned}\]where we used that \(\{\phi = 0\} \times \{0\}\) has measure zero, that the sets \(A_{i} \times [0,a_{i}]\) are pairwise disjoint because the \(A_{i}\) are, and the formula \(m(A_{i} \times [0,a_{i}]) = a_{i}\, m(A_{i})\).
Theorem (Monotonicity of the integral with respect to the integrand and the domain)
Let \(f, g : E \to [0,\infty)\) be measurable.
-
If \(0 \leq g \leq f\), then
\[\int_{E} g(x) \, dx \leq \int_{E} f(x) \, dx.\] -
If \(E_{1} \subseteq E_{2} \subseteq E\) are measurable, then
\[\int_{E_{1}} f(x) \, dx \leq \int_{E_{2}} f(x) \, dx.\]
Proof: Since \(0 \leq g \leq f\), we have that
\[\{ (x,y) :\ x \in E,\ 0 \leq y \leq g(x) \} \subseteq \{ (x,y) :\ x \in E,\ 0 \leq y \leq f(x) \},\]hence \(R(g,E) \subseteq R(f,E)\), and part (i) follows from the monotonicity of the measure.
For the second part we shall first prove that
\[\int_{E_{1}} f(x) \, dx = \int_{E} \mathbf{1}_{E_{1}} f(x) \, dx\]for \(E_{1} \subseteq E\) measurable. Note that
\[\{ (x,y) :\ x \in E,\ 0 \leq y \leq \mathbf{1}_{E_{1}}(x) f(x) \} = \big( (E \setminus E_{1}) \times \{0\} \big) \cup \{ (x,y) :\ x \in E_{1},\ 0 \leq y \leq f(x) \},\]and since \((E \setminus E_{1}) \times \{0\}\) has measure zero, then
\[m\big( R(\mathbf{1}_{E_{1}} f, E) \big) = m\big( R(f,E_{1}) \big).\]Now, if \(E_{1} \subseteq E_{2}\) are measurable, then
\[\mathbf{1}_{E_{1}} f \leq \mathbf{1}_{E_{2}} f,\]and by part (i) we obtain
\[\int_{E} \mathbf{1}_{E_{1}} f(x) \, dx \leq \int_{E} \mathbf{1}_{E_{2}} f(x) \, dx.\]We conclude that
\[\int_{E_{1}} f(x) \, dx \leq \int_{E_{2}} f(x) \, dx.\]