Let \((X, \mathcal M, \mu)\) be a measure space. If \(f \in L^1(X)\) and \(\varepsilon>0\), there is an integrable simple function \(\phi=\sum_{j=1}^na_j \mathbf{1}_{{A_j}}\) such that \[\int_X |f-\phi|d\mu<\varepsilon.\] That is, the integrable simple functions are dense in \(L^1\) in the \(L^1\) metric.
Furthermore, if \(\mu\) is a Lebesgue–Stieltjes measure on \(\mathbb R\) (or Lebesgue measure on \(\mathbb R^d\)), the sets \(A_j\) in the definition of \(\phi\) can be taken to be finite unions of open intervals (open rectangles), and also there is a continuous function \(g\) that vanishes outside a bounded interval (bounded rectangle) such that \[\int_X |f-g|d\mu<\varepsilon.\]
Proof. We fix \(f \in L^1(X)\) and take a sequence \((\phi_n)_{n \in \mathbb{N}}\) of measurable simple functions \(|\phi_1| \leq |\phi_2| \leq \ldots \leq |f|\) such that \(\phi_n \ _{\overrightarrow{n \to \infty}} \ f\) a.e.. Then \[\lim_{n \to \infty}|\phi_n-f|=0, \quad \text{ and } \quad |\phi_n-f| \leq 2|f|.\] Thus by the (DCT), we obtain \[\lim_{n \to \infty}\int_X |\phi_n-f|d\mu=0.\] If \(\phi_k=\sum_{j=1}^na_{j} \mathbf{1}_{{A_{j}}}\), where the \(A_{j}\)’s are disjoint and the \(a_{j}\)’s are nonzero, we note that \[\mu(A_j)=a_j^{-1}\int_{A_j}|\phi_k|d\mu\le a_j^{-1}\int_X|f|d\mu<\infty.\]
If \(E, F\in\mathcal M\) note that \(\mu(E\triangle F)=\int_X|\mathbf{1}_{{E}}-\mathbf{1}_{{F}}|d\mu\) and recall that:
If \(E\in \mathcal L(\mathbb R^d)\) and \(\lambda_d(E)<\infty\), then for every \(\varepsilon>0\) there is a set \(A\) that is a finite union of open rectangles such that \(\lambda_d(E\triangle A)<\varepsilon.\) The same result is true if \(\mu\) is a Lebesgue–Stieltjes measure on \(\mathbb R\).
If \(\mu\) is a Lebesgue–Stieltjes measure (or Lebesgue measure of \(\mathbb R^d\)) we can approximate \(\mathbf{1}_{{A_j}}\) arbitrarily closely in the \(L^1\) metric by finite sums of \(\mathbf{1}_{{R_k}}\), where the \(R_k\)’s are open intervals (or open rectangles).
Finally, if \(R_k=\prod_{i=1}^d(a_i, b_i)\) we can approximate \(\mathbf{1}_{{R_k}}\) in the \(L^1\) metric by continuous functions that vanish outside \(\prod_{i=1}^d(a_i, b_i)\).
For instance, if \(d=1\), given \(\varepsilon>0\) take a continuous picewise linear function \(g\) such that \(\mathbf{1}_{{[a+\varepsilon, b-\varepsilon]}}\le g\le \mathbf{1}_{{(a, b)}}\). $$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a measure space and suppose that \(f:X \times [a,b] \to \mathbb{C}\) (\(-\infty<a<b<\infty\)) and \(f(\cdot,t)\in L^1(X)\) for each \(t \in [a,b]\). Let \[F(t)=\int_X f(x,t)\,d\mu(x).\]
Suppose that there exists \(g \in L^1(X)\) such that \[|f(x,t)| \leq |g(x)| \quad \text{ for all } \quad (x, t)\in X \times [a,b].\] If \[\lim_{t \to t_0}f(x,t)=f(x,t_0) \quad \text{ for every } \quad x \in X,\] then \[\lim_{t \to t_0}F(t)=F(t_0).\]
Proof. Let \[f_n(x)=f(x,t_n),\] where \((t_n)_{n \in \mathbb{N}} \subseteq [a,b]\) is converging to \(t_0 \in [a,b]\). Note that \[|f_n(x)| \leq |g(x)| \in L^1(\mu),\] and \[\lim_{n\to \infty}f_n(x)=f(x,t_0).\] Thus by the (DCT) we obtain \[\lim_{n \to \infty}F(t_n)=\lim_{n \to \infty}\int_X f_n(x)d\mu(x) =\int_X \lim_{n \to \infty}f_n(x)d\mu(x)=\int_X f(x,t_0)d\mu(x),\] which proves that \[\lim_{n \to \infty}F(t_n)=F(t_0)\] as desired. $$\tag*{$\blacksquare$}$$
Let \((X, \mathcal M, \mu)\) be a measure space and suppose that \(f:X \times [a,b] \to \mathbb{C}\) (\(-\infty<a<b<\infty\)) and \(f(\cdot,t)\in L^1(X)\) for each \(t \in [a,b]\). Let \[F(t)=\int_X f(x,t)\,d\mu(x).\]
Suppose that \(\frac{\partial f}{\partial t}\) exists and there is \(g \in L^1(X)\) such that \[\left|\frac{\partial f}{\partial t}(x,t)\right| \leq |g(x)| \quad \text{ for all } \quad (x, t)\in X \times [a,b].\] Then \(F\) is differentiable and \[F'(t)=\frac{\partial F}{\partial t}(x)=\int_X \frac{\partial f}{\partial t}(x,t)d\mu(x).\]
Proof. Suppose that \((t_n)_{n \in \mathbb{N}} \subseteq [a,b]\) is converging to \(t_0 \in [a,b]\), then we may write \[\frac{\partial f}{\partial t}(x,t_0)=\lim_{n \to \infty}h_n(x),\] where \[h_n(x)=\frac{f(x,t_n)-f(x,t_0)}{t_n-t_0}.\] It follows that \(\frac{\partial f}{\partial t}\) is measurable and by the mean-value theorem one has \[|h_n(x)| \leq \sup_{t \in [a,b]}\left|\frac{\partial f}{\partial t}(x,t)\right| \leq |g(x)|,\] so by the (DCT) we conclude \[F'(t_0)=\lim_{n \to \infty}\frac{F(t_n)-F(t_0)}{t_n-t_0}=\lim_{n \to \infty}\int_X h_n(x)d\mu(x)=\int_X \frac{\partial f}{\partial t}(x,t_0)d\mu(x)\] as claimed. $$\tag*{$\blacksquare$}$$
Let \([a, b]\) be a given interval. By a partition \(P\) of \([a, b]\) we mean a finite set of points \(a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b.\)
Suppose that \(f:[a,b]\to \mathbb R\) is a bounded function. Corresponding to each partition \(P\) of \([a, b]\) we put \[\begin{gathered} m_i=\inf_{x \in [x_{i-1},x_i]}f(x), \quad \text{ and } \quad M_i=\sup_{x \in [x_{i-1},x_i]}f(x). \end{gathered}\]
Upper and lower Riemann sums We define \[U(P,f)=\sum_{i=1}^{n}M_i(x_i-x_{i-1}), \quad \text{ and } \quad L(P,f)=\sum_{i=1}^{n}m_i(x_i-x_{i-1}).\]
We always have that \(L(P,f)\le U(P,f)\).
Definition. We define the upper and lower Riemann integrals of \(f\) over \([a, b]\) to be \[\begin{aligned} L_a^b(f)=\sup_P L(P,f), \quad \text{ and } \quad U_a^b(f)=\inf_P U(P,f), \end{aligned}\] where the inf and the sup are taken over all partitions \(P\) of \([a,b]\).
Riemann integral of \(f\) over \([a, b]\) If the upper and lower integrals are equal, we say that \(f:[a, b]\to \mathbb R\) is Riemann integrable on \([a, b]\), we write \(f \in \mathcal{R}([a, b])\) and we denote the common value (which is called Riemann integral of \(f\) over \([a, b]\)) by \[\int_a^bf(x)dx=L_a^b(f)= U_a^b(f).\]
Fact The upper and lower integrals are defined for every bounded function.
Proof. Let \[\begin{aligned} m &=\inf_{x \in [a,b]} f(x),\\ M& =\sup_{x \in [a,b]}f(x). \end{aligned}\] Then \[m \leq f(x) \leq M \quad \text{ for all }\quad x \in [a,b].\] Therefore \[m(b-a) \leq L(P,f) \leq U(P,f) \leq M(b-a)\] for every partition \(P\).$$\tag*{$\blacksquare$}$$
There is a bounded function \(f\in L^1([0, 1])\) which is not Riemann integrable.
Proof. Let us define \(f\) on \([0,1]\) to be \[f(x)=\begin{cases} 1 \text{ if }x \in \mathbb{Q},\\ 0 \text{ if }x\not\in \mathbb{Q}. \end{cases}\] We immediately see that \(f\in L^1(\mathbb R)\). Let us recall
Fact (*) In any interval \([c,d]\) such that \(c<d\) there is a rational and irrational number.
By the fact \({\color{purple}(*)}\), for every partition \(P\) of \([0, 1]\), we have
\[U(P,f)=\sum_{i=1}^{n}M_i(x_i-x_{i-1})=\sum_{i=1}^n1\cdot (x_i-x_{i-1}) =1,\] \[L(P,f)=\sum_{i=1}^{n}m_i(x_i-x_{i-1})=\sum_{i=1}^n 0\cdot (x_i-x_{i-1})=0.\]
Therefore
\[L_a^b(f)=\sup_P L(P,f)=0, \quad \text{ and } \quad U_a^b(f)=\inf_P U(P,f)=1.\]
Hence \({\color{red}L_a^b(f) \neq U_a^b(f)}\) and \(f\) is not Riemann integrable.$$\tag*{$\blacksquare$}$$
Proof. We prove the first statement.
Suppose first that \(P^{*}\) contains just one point more than \(P\). Let this extra point be \(x^{*}\), and suppose \[x_{i-1} \leq x^{*} \leq x_i \quad \text{ for some }\quad i \in \{1,2,\ldots,n\}.\]
Let \[\begin{gathered} m_i=\inf_{x \in [x_{i-1},x_i]}f(x),\\ w_1=\inf_{x \in {\color{red}[x_{i-1},x^{*}]}}f(x), \quad \text{ and } \quad w_2=\inf_{x \in {\color{blue}[x^{*},x_i]}}f(x) \end{gathered}\]
Then \(w_1\ge m_i\) and \(w_2\ge m_i\) and consequently \[\begin{gathered} L(P^{*},f)-L(P,f)={\color{red}w_1(x^{*}-x_{i-1})}+{\color{blue}w_2(x_i-x^*)} -{\color{purple}m_i(x_i-x_{i-1})}\\ ={(w_1-m_i)(x^{*}-x_{i-1})}+{(w_2-m_i)(x_i-x^*)} \geq 0. \end{gathered}\]
Finally, if \(P^*\) contains \({\color{brown}k}\) points more than \(P\), we repeat this reasoning \({\color{brown}k}\) times. The proof of the second statement is analogous.$$\tag*{$\blacksquare$}$$
Claim (*) For two partitions \(P_1, P_2\) of an interval \([a, b]\) one has \[L(P_1,f) \leq U(P_2,f).\]
Proof. Let \(P^{*}=P_1\cup P_2\) be the common refinement of two partitions \(P_1\) and \(P_2\). By the previous theorem \[{\color{red}L(P_1,f) \leq L(P^*,f)} \leq {\color{blue}U(P^*,f) \leq U(P_2,f)}.\]
Proof. By the Claim (*) for two partitions \(P_1, P_2\) of an interval \([a, b]\) one has \[L(P_1,f) \leq U(P_2,f).\] Then \[L_a^b(f)=\sup_{P_1} L(P_1,f)\le \inf_{P_2} U(P_2,f)= U_a^b(f)\] This completes the proof of the theorem.$$\tag*{$\blacksquare$}$$
For every \(\varepsilon>0\) there is a partition \(P\) of \([a, b]\) such that \[\begin{aligned} \qquad\qquad \qquad U(P,f)-L(P,f)<\varepsilon. \qquad\qquad \qquad {\color{purple} (\mathcal R)} \end{aligned}\]
Proof. By the previous theorem, for every partition \(P\) we have \[L(P,f) \leq L_a^b(f) \leq U_a^b(f) \leq U(P,f).\] Thus the condition (\(\mathcal R\)) implies \[0\le U_a^b(f) - L_a^b(f)\le U(P,f)-L(P,f)<\varepsilon.\] Since \(\varepsilon>0\) is arbitrary \(U_a^b(f) = L_a^b(f)\), hence \(f \in \mathcal{R}([a, b])\).
Conversely, suppose that \(f \in \mathcal{R}([a, b])\). Then for every \(\varepsilon>0\) there are partitions \({\color{red}P_1}\) and \({\color{blue}P_2}\) such that \[\begin{aligned} {\color{red}L_a^b(f)-L(P_1,f)<\frac{\varepsilon}{2}} \quad \text{ and } \quad {\color{blue}U(P_2,f)-U_a^b(f)<\frac{\varepsilon}{2}}. \end{aligned}\]
We choose \(P\) to be the common refinement of \({\color{red}P_1}\) and \({\color{blue}P_2}\). Then \[\begin{gathered} U(P,f) \leq U(P_2,f)\\ \leq U_a^b(f)+\frac{\varepsilon}{2}= {\int_a^b}f(x)dx+\frac{\varepsilon}{2}= L_a^b(f)+\frac{\varepsilon}{2}\\ \leq L(P_1,f)+\varepsilon \leq L(P,f)+\varepsilon. \end{gathered}\] This proves condition (\(\mathcal R\)) and completes the proof of the theorem. $$\tag*{$\blacksquare$}$$
Proof. Note that \(f(s_i),f(t_i)\) lies in \([m_i,M_i]\), hence by the triangle inequality \(|f(t_i)-f(s_i)| \leq \underbrace{M_i-m_i}_{\text{length}}.\) Hence \[\begin{gathered} \sum_{i=1}^{n}|f(s_i)-f(t_i)|(x_i-x_{i-1}) \leq \sum_{i=1}^{n}(M_i-m_i)(x_i-x_{i-1})\\ =U(P, f) -L(P, f) <\varepsilon. \end{gathered}\] This completes the proof. $$\tag*{$\blacksquare$}$$
Proof. It is enough to note that \[\begin{aligned} L(P,f) &\leq \sum_{i=1}^nf(t_i)(x_i-x_{i-1}) \leq U(P,f),\\ L(P,f) &\leq \int_a^b f(x) dx \leq U(P,f).\quad {\blacksquare} \end{aligned}\]
Let \(f:[a,b] \to \mathbb{R}\) be a bounded function . If \(f\) is Riemann integrable then \(f\) is Lebesgue measurable and integrable on \([a,b]\) and \[{\color{blue}\underbrace{\int_a^b f(x)dx}_{\text{Riemann}}}={\color{red}\underbrace{\int_{[a,b]}f(x)d\lambda(x)}_{\text{Lebesgue}}}.\]
Proof. Assume that \(f\in \mathcal R([a, b])\). For each partition \(P=(t_j)_{j=0}^n\) consider \[G_P=\sum_{j=1}^n M_j\mathbf{1}_{{[t_{j-1},t_j]}}, \quad \text{ and } \quad g_P=\sum_{j=1}^n m_j\mathbf{1}_{{[t_{j-1},t_j]}}.\] Then \[U(P, f)=\int_{[a, b]} G_P(x)d\lambda(x), \quad \text{ and } \quad L(P, f)=\int_{[a, b]} g_P(x)d\lambda(x).\]
By the Riemann integrability there is a sequence \((P_k)_{k \in \mathbb{N}}\) of partitions of \([a,b]\) such that \(P_k=\{t_{k, 0},\ldots, t_{k, n_k}\}\) such that \[a=t_{k, 0}< t_{k, 1}<\ldots< t_{k, n_k-1}< t_{k, n_k}=b\] and \(P_{k}\subset P_{k+1}\) for any \(k\in\mathbb N\). Moreover, \[\lim_{k\to \infty}\max_{1\le \:j \le n_k} (t_{k, j}-t_{k, j-1})=0,\] and \[\lim_{k \to \infty}U(P_k, f)=\lim_{k \to \infty}L(P_k, f)=\int_a^b f(x)dx.\] We also see that \(g_{P_k}\) increases and \(G_{P_k}\) decreases as \(k \to \infty\), thus we may write \[G=\lim_{k \to \infty}G_{P_k}, \quad \text{ and }\quad g=\lim_{k \to \infty}g_{P_k}.\]
Then \(g \leq f \leq G\) and by the (DCT), we conclude \[\int_{[a, b]} G(x)d\lambda(x)=\int_{[a, b]} g(x)d\lambda(x)=\int_a^b f(x)dx.\]
Hence \[\int_{[a, b]} (G(x)-g(x))d\lambda(x)=0,\] so \(G=g\) a.e., and consequently \(G=g=f\) a.e..
But \(G\) is measurable (as a limit of a sequence of simple functions) and \(\lambda\) is complete, thus \(f\) is measurable and \[\qquad{\color{blue}\underbrace{\int_a^b f(x)\,dx}_{\text{Riemann}}}=\int_{[a, b]} G(x)d\lambda(x) ={\color{red}\underbrace{\int_{[a, b]}f(x)d\lambda(x)}_{\text{Lebesgue}}}. \qquad \tag*{$\blacksquare$}\]
A bounded function \(f\in \mathcal R([a, b])\) iff \(f\) is a.e. continuous.
Proof of (\(\Longrightarrow\)). Assume that \(f\in \mathcal R([a, b])\). For \(x\in [a, b]\) define \[\begin{aligned} g(x)&=\sup_{\delta>0} \inf_{y\in [a, b]\cap B(x, \delta)}f(y),\\ h(x)&=\inf_{\delta>0} \sup_{y\in [a, b]\cap B(x, \delta)}f(y), \end{aligned}\] where \(B(x, \delta)=(x-\delta, x+\delta)\). It is not difficult to see that \(g\le f\le h\) and \(f\) is continuous at \(x\in [a, b]\) iff \(g(x)=h(x)\). Moreover, for any partition \(P\) of \([a, b]\) we have \[\begin{aligned} L(P, g)&\le L(P, f)\le L(P, h),\\ U(P, g)&\le U(P, f)\le U(P, h). \end{aligned}\]
In fact, we have \(U(P, f)= U(P, h)\) and \(L(P, g)= L(P, f)\), since for any open interval \((c, d)\subset [a, b]\) we have \[\begin{aligned} \inf_{x\in(c, d)}g(x)=\inf_{x\in(c, d)}f(x), \quad \text{ and } \quad \sup_{x\in(c, d)}h(x)=\sup_{x\in(c, d)}f(x). \end{aligned}\] Indeed, for any \(z\in (c, d)\) and \(B(z, \delta)\subseteq (c, d)\) we have \[\begin{aligned} \inf_{x\in(c, d)}g(x)\le \inf_{x\in(c, d)}f(x)\le \inf_{y\in B(z, \delta)}f(y)\le g(z). \end{aligned}\] Since \(z\in (c, d)\) is arbitrary we obtain \(\inf_{x\in(c, d)}g(x)=\inf_{x\in(c, d)}f(x)\). Similarly, we show that \(\sup_{x\in(c, d)}h(x)=\sup_{x\in(c, d)}f(x)\). It follows that \[\begin{aligned} &L_a^b(f)= L_a^b(g)\le U_a^b(g)\le U_a^b(f),\\ &L_a^b(f)\le L_a^b(h)\le U_a^b(h)= U_a^b(f), \end{aligned}\] and consequently \(g, h\in \mathcal R([a, b])\), since \(f\in \mathcal R([a, b])\).
Since \(g\le h\) and \(\int_{[a, b]}(h-g)d\lambda=0\) we obtain that \(g=h\) a.e., and consequently \(g=h=f\) a.e., thus \(f\) is continuous a.e. as we claimed.
Proof of (\(\Longleftarrow\)). Assume that \(f\) is continuous a.e., and for \(n\in \mathbb N\) let \(P_n\) be the dissection of \([a, b]\) into \(2^n\) equal points. Set \[h_n(x)=\sup_{y\in (c, d)}f(y), \quad \text{ and } \quad g_n(x)=\inf_{y\in (c, d)}f(y),\] where \((c, d)\) is an open interval containing \(x\), whose endpoints are two consecutive elements from \(P_n\), and set \(h_n(x)=g_n(x)=f(x)\) if \(x\in P_n\).
Then we see that \((g_n)_{n\in\mathbb N}\) and \((h_n)_{n\in\mathbb N}\) are respectively, increasing and decreasing sequences of functions, each function constant on each of a finite family of intervals covering \([a, b]\), and \[U(P_n, f)=\int_{[a, b]} h_n(x)d\lambda(x), \quad \text{ and } \quad L(P_n, f)=\int_{[a, b]} g_n(x)d\lambda(x).\]
We also see that \[\lim_{n\to \infty}g_n(x)=\lim_{n\to \infty}h_n(x)=f(x)\] at any point \(x\in [a, b]\) at which \(f\) is continuous. By the (DCT), we conclude that \[\lim_{n\to \infty}\int_{[a, b]}g_n(x)d\lambda(x)= \int_{[a, b]}f(x)d\lambda(x)= \lim_{n\to \infty}\int_{[a, b]}h_n(x)d\lambda(x),\] but this means that \[L_a^b(f)\ge \int_{[a, b]}f(x)d\lambda(x)\ge U_a^b(f),\] so these are all equal and \(f\) is Riemann integrable. $$\tag*{$\blacksquare$}$$