Proof. The sequence of functions \begin{align*} f_n(\omega) = \begin{cases} -n & -\infty \leq f(\omega)\leq -n\\ -(k-1)2^{-n} & -k 2^{-n}< f(\omega)\leq -(k-1)2^{-n}, \quad 1\leq k\leq n2^n\\ (k-1)2^{-n} & (k-1)2^{-n}\leq f(\omega) < k 2^{-n}, \quad 1\leq k \leq n2^n\\ n & n\leq f(\omega)\leq +\infty \end{cases} \end{align*} has the necessary properties (see the following example for an illustration).

X = seq(-2, 2, length.out = 200)
f1 = rep(0, length.out = 200)
f1[X^2 > 1/2 & X^2 < 1] = 1/2
f1[X^2 > 1] = 1
qplot(X, X^2, xlab = "$x$", ylab = "$f(x)=x^2$", geom = "line",
    main = "Approximating a Quadratic Function by a Simple Function ($n=1$)") +
    geom_area(aes(x = X, y = f1))

f2 = rep(0, length.out = 200)
f2[X^2 > 1/4 & X^2 < 2/4] = 1/4
f2[X^2 > 2/4 & X^2 < 3/4] = 2/4
f2[X^2 > 3/4 & X^2 < 4/4] = 3/4
f2[X^2 > 4/4 & X^2 < 5/4] = 4/4
f2[X^2 > 5/4 & X^2 < 6/4] = 5/4
f2[X^2 > 6/4 & X^2 < 7/4] = 6/4
f2[X^2 > 7/4 & X^2 < 8/4] = 7/4
f2[X^2 > 8/4] = 2
qplot(X, X^2, xlab = "$x$", ylab = "$f(x)=x^2$", geom = "line",
    main = "Approximating a Quadratic Function by a Simple Function ($n=2$)") +
    geom_area(aes(x = X, y = f2))

Proof. These properties follow directly from Definitions F.3.2-F.3.3. The first property follows from the fact that the supremum in the definition of $\int_E f\, d\mu$ is less than or equal to the supremum in the definition of $\int_E g\, d\mu$. The second property holds since the supremum in the definition of $\int_A f\,d\mu$ is less than or equal than the supremum in the definition of $\int_B f\,d\mu$. The third property follows from the fact that the supremum in the definition of $\int_E cf\,d\mu$ is $c$ times the supremum in the definition of $\int_E f\,d\mu$. The fourth property follows from the fact that the supremum in the definition of $\int_E f\,d\mu$ can only be the zero function. The fifth property follows since the integrals of all simple functions over $E$ are zero. The sixth property follows directly from Definitions F.3.2-F.3.3.

Proof. We assume that $s=\sum_{i=1}^k \alpha_i I_{A_i}$ and $t=\sum_{i=1}^l \beta_i I_{B_i}$. For all sets $E\in\mathcal{F}$, we have $\phi(E)\geq 0$ and $\phi(\emptyset)=0$. If $E_n, n\in\mathbb{N}$ is a sequence disjoint sets in $\mathcal{F}$ with $\cup_n E_n=E$ then \begin{align*} \phi(E) &= \int_E \sum_{i=1}^k \alpha_i I_{A_i}\, d\mu = \sum_{i=1}^k \alpha_i \mu(A_i\cap E) = \sum_{i=1}^k \sum_{n\in\mathbb{N}}\alpha_i \mu(A_i\cap E_n)\\ &= \sum_{n\in\mathbb{N}} \phi(E_n), \end{align*} proving that $\phi$ is a measure.

For all $E_{ij}=A_i\cap B_j$, $i=1,\ldots,k$, $j=1,\ldots,l$ \begin{align*} \int_{E_{ij}} (s+t)\, d\mu = (\alpha_i+\beta_j) \mu(E_{ij}) = \int_{E_{ij}} s\, d\mu + \int_{E_{ij}} t\, d\mu. \end{align*} Since $\phi$ is a measure, it is countable additive and so the above equation implies $\int_{\Omega} (s+t)\, d\mu = \int_{\Omega} s\, d\mu + \int_{\Omega} t\, d\mu.$

Proof. The function $f=\lim f_n$ is measurable by Proposition E.6.4, and since $\int f_n\,d\mu \nearrow$, we have $\int f_n\,d\mu\nearrow \alpha$ for some $\alpha\in[0,+\infty]$ (see Proposition B.2.1). Since $f_n \leq \lim f_n$, we have $\int f_n\,d\mu \leq \int_{\Omega} \lim_{n\to\infty} f_n \, d\mu$.

It remains to show the reverse inequality $\int f_n\,d\mu \geq \int_{\Omega} \lim_{n\to\infty} f_n \, d\mu$. For any non-negative simple measurable function $s$ such that $s\leq \lim f_n$ and any $c\in (0,1)$ we define $E_n=\{x:f_n(x)\geq cs(x)\}, n\in\mathbb{N}$. Note that $E_n\in\mathcal{F}$, $E_n\nearrow$, and $\Omega=\cup_n E_n$. By definition of $E_n$, we have for all $n\in\mathbb{N}$ \begin{align} \int_{\Omega} f_n \, d\mu \geq \int_{E_n} f_n\,d\mu \geq c\int_{E_n} s\, d\mu. \end{align} We let $n\to\infty$ and recall that $\int_{E_n} s\, d\mu$ is a measure (Proposition F.3.3) $\phi(E_n)$, which converges by Proposition E.2.1 to $\phi(\Omega)$. This implies that \[\lim \int_{\Omega} f_n \, d\mu \geq \lim c\int_{E_n} s\, d\mu = c \int_{\Omega} s\, d\mu.\] Since this holds for any $c\in (0,1)$ we also have \[\lim \int_{\Omega} f_n \, d\mu \geq \int_{\Omega} s\, d\mu.\] Finally, since this holds for all non-negative simple measurable function $s$ such that $s\leq \lim f_n$, it also holds for their supremum, which defines the integral $\int_{\Omega} \lim f_n\, d\mu$.

Proof. We first prove this result for a sum of two functions $f_1+f_2$. Let $s_k$ and $t_k$ be the sequence of measurable functions that converge to $f_1$ and $f_2$ as in Proposition F.3.1. Then the sequence of simple functions $s_k+t_k$ converge to $f_1+f_2$, and using the Monotone Convergence Theorem above together with Proposition F.3.3, we have \begin{align*} \int_{\Omega} f_1+f_2\,d\mu &= \int_{\Omega} \lim_k (s_k+t_k)\,d\mu = \lim_k \int_{\Omega} s_k+ t_k \,d\mu\\ &=\lim_k \int_{\Omega} s_k\, d\mu + \lim_k \int_{\Omega} t_k \,d\mu = \int_{\Omega} f_1\,d\mu + \int_{\Omega} f_2\,d\mu. \end{align*} By induction, we establish the result for a sum of $N$ functions. Defining $g_N=\sum_{n=1}^N f_n$, we have that $g_N\nearrow \sum_{n=1}^{\infty} f_n$. Applying the Monotone Convergence Theorem again yields \begin{align*} \int_{\Omega} \sum_{n=1}^{\infty} f_n\, d\mu &= \int_{\Omega} \lim_{N\to\infty} g_N \, d\mu \\ &= \lim_{N\to \infty} \int_{\Omega} g_N \, d\mu \\ &= \lim_{N\to\infty} \sum_{n=1}^{N} \int_{\Omega} f_n\, d\mu\\ &=\sum_{n=1}^{\infty} \int_{\Omega} f_n\, d\mu. \end{align*}

Proof. It is easy to see that $\phi(E)\geq 0$ for all $E\in\mathcal{F}$ and $\phi(\emptyset)=0$. We next show that $\phi$ is countable additive. Let $E_n, n\in\mathbb{N}$ be a sequence of disjoint sets whose union equals $E$. Corollary F.3.1 implies that \begin{align} \tag{*} \phi(E) = \int_{\Omega} f I_E\,d\mu = \int_{\Omega} \sum_{n=1}^{\infty} f I_{E_n} \,d\mu = \sum_{n=1}^{\infty} \int f I_{E_n}\,d\mu = \sum_{n=1}^{\infty} \phi(E_n). \end{align} The second statement above follows from the first statement whenever $g$ is an indicator function $g(x)=cI_A(x)$. Similar arguments show that the second statement follows from the first statement whenever $g$ is a simple function. The case of a general function $g$ follows from approximating $g$ by a sequence of simple functions $0\leq s_1\leq s_2\leq\cdots$ (Proposition F.3.1) and using the Monotone Convergence Theorem twice \[ \int_{\Omega} g \, d\phi = \int_{\Omega} \lim_n s_n \, d\phi = \lim_n \int_{\Omega} s_n\,d \phi = \lim_n \int_{\Omega} s_n f\,d\mu = \int_{\Omega} g f\,d\mu.\]

Proof. Defining the function $g_k(x)=\inf\{f_i(x): i\geq k\}$, we have $g_k\leq f_k$ and therefore for all $k\in\mathbb{N}$, \[\int_{\Omega} g_k\,d\mu\leq \int_{\Omega} f_k \,d\mu.\] Since $0\leq g_1\leq g_2\leq\cdots$, and $g_k\to\liminf_n f_n$, we can apply the Monotone convergence theorem to the left hand side above to get \[ \int_{\Omega} \liminf_n f_n\,d\mu = \lim_k \int_{\Omega} g_k\,d\mu =\liminf_k \int_{\Omega} g_k\,d\mu \leq \liminf_k \int_{\Omega} f_k \,d\mu.\]

Proof. The left hand side equals $\int f^+\,d\mu - \int f^-\,d\mu$, a difference of two non-negative quantities. This is clearly smaller than the right hand side, which equals $\int |f^+|\,d\mu + \int |f^-|\,d\mu$.

Proof. We decompose the proof into the case $\int_{\Omega} f+g\,d\mu = \int_{\Omega} f\,d\mu + \int_{\Omega} g\, d\mu$ and the case $\int_{\Omega} \alpha f \,d\mu = \alpha \int_{\Omega} f\,d\mu$. Both cases together imply the proposition above.

In the first case, if $\alpha > 0$ the result follows from applying Corollary F.3.1 to the positive-negative decomposition \[\int \alpha f\, d\mu = \int (\alpha f)^+ \, d\mu - \int (\alpha f)^- \, d\mu = \alpha \int f^+ \, d\mu - \alpha \int f^- \, d\mu = \alpha \int f\,d\mu.\] Similarly, if $\alpha<0$ \begin{align*} \int \alpha f\, d\mu &= \int (\alpha f)^+ \, d\mu - \int (\alpha f)^- \, d\mu \\ &= \int (-\alpha f^-) \, d\mu - \int (-\alpha f^+) \, d\mu \\ &= -\alpha \int f^- \, d\mu - (-\alpha) \int f^+ \, d\mu = \alpha \int f\,d\mu. \end{align*} In the second case, the result follows from applying Corollary F.3.1 to the positive-negative decomposition of $f+g$ \begin{align*} \int f+g\,d\mu &= \int f^+ - f^- + g^+ - g^-\,d\mu \\ &= \int f^+\,d\mu - \int f^-\,d\mu + \int g^+\,d\mu - \int g^- \, d\mu \\ &= \int f\,d\mu + \int g\,d\mu. \end{align*}

Proof. We denote $f=\lim f_n$ and observe that $f_n$ is integrable since $|f_n|\leq g$ and $g$ is integrable. Applying Fatou's lemma to the sequence of non-negative functions $2g-|f_n-f|\geq 0$ we get \begin{align*} \int 2g \,d\mu &\leq \liminf_{n\to\infty} \int (2g-|f_n-f|)\,d\mu = \int 2g\,d\mu + \liminf_{n\to\infty} - \int |f_n-f|\,d\mu\\ & = \int 2g\,d\mu - \limsup_{n\to\infty} \int |f_n-f|\,d\mu, \end{align*} which implies \begin{align*} 0 \geq \limsup_{n\to\infty} \int |f_n-f|\,d\mu. \end{align*} Since $\int |f_n-f|\,d\mu$ is a sequence of non-negative values, the first result holds. The second result follows from applying Proposition F.3.7 to $f_n-f$: \begin{align*} 0 &= \lim_{n\to\infty} \int_{n\to\infty} |f_n-f|\,d\mu \\ &\geq \lim_{n\to\infty} \Big|\int f_n-f \,d\mu\Big | \\ &= \lim_{n\to\infty} \Big| \int f_n\,d\mu - \int f \,d\mu \Big |. \end{align*}

Proof. The result follows immediately from the second statement of the dominated convergence theorem above.

Proof. Defining $E_n$ to be the subset of $\Omega$ on which $f>1/n$, we have $f\geq \sum_{n\in\mathbb{N}} (1/n) I_{E_n}$ and \[ 0=\int f\,d\mu \geq \int \sum_{n\in\mathbb{N}} (1/n) I_{E_n}\,d\mu = \sum_{n\in\mathbb{N}} (1/n) \mu(E_n).\] It follows that $\mu(E_n)=0$ for all $n\in\mathbb{N}$, implying that $\mu(\cup_{n\in\mathbb{N}}E_n) \leq \sum_{n\in\mathbb{N}}\mu(E_n)=0$ and therefore $f=0$ except on a set of measure zero.

Proof. We use notations below from the section concerning Riemann integrals. To distinguish between the Riemann and the Lebesgue integrals we denote the former as $\int f\,dx$ and the latter as $\int f\,d\mu$ ($\mu$ here corresponds to the Lebesgue measure). All integrals in the proof below are over the interval $[a,b]$.

We construct a sequence of partitions $P_n, n\in\mathbb{N}$ such that $P_k$ is a refinement of $P_{k-1}$, the distance between adjacent points in $P_k$ is less than $1/k$, and \begin{align*} \lim_{k\to\infty} U(P_k,f)&=\overline{\int f\,dx}\\ \lim_{k\to\infty} L(P_k,f)&=\underline{\int f\,dx}. \end{align*} (This is possible since the lower and upper Riemann integrals are defined using the limits above.) Defining the following two simple functions \begin{align} L(x) &= \begin{cases} m_i & x\in\Delta_i\\ f(a) & x=a \\ f(b) & x=b\end{cases}\\ U(x) &= \begin{cases} M_i & x\in\Delta_i\\ f(a) & x=a \\ f(b) & x=b\end{cases}. \end{align} we have \begin{align*} L(P_k,f)&=\int L_k\,d\mu \\ U(P_k,f)&=\int U_k\,d\mu \\ L_1\leq L_2\leq \cdots\leq &f\leq \cdots\leq U_2\leq U_l \quad \text{(since } P_k \text{ refines } P_{k-1}). \end{align*} The last equation above implies that $L_k$ and $U_k$ converge to limit functions $L(x)=\lim_{n\to\infty} L_k(x)$ and $U(x)=\lim_{n\to\infty} U_k(x)$ and that $L\leq f\leq U$. Using the Monotone Convergence Theorem, we have \[ \int L\,d\mu = \underline{\int f\,dx},\qquad \int U\,d\mu = \overline{\int f\,dx}.\]

Note that the above derivations assume only that $f$ is bounded. The function $f$ is Riemann integrable, if and only if the lower and upper integrals agree, which occurs if and only if $L=U$ a.e. (using the fact that $U-L\geq 0$ and Proposition F.3.10) or equivalently $L=f=U$. This proves that if $f$ is a bounded Riemann integrable function, it is also Lebesgue integrable with respect to the Lebesgue measure and the two integrals agree.

If $x$ is not equal to one of the points in the partition $P_k$, then $L(x)=U(x)$ if and only if $m_i=M_i$, which occurs if and only if $f$ is continuous at $x$. Recalling that $f$ is Riemann integrable if and only if $L=U$ a.e. (shown above), we have that $f$ is Riemann integrable if and only if $f$ is continuous a.e..

Proof. If $f=I_A$ then $f\circ T=I_{T^{-1}A}$ and (**) holds by definition of the Lebesgue integral and the transformed measure. The linearity of the integral implies that (**) holds also for non-negative simple function. The change of variable formula (**) holds for non-negative function by constructing a sequence of simple functions $s_n, n\in\mathbb{N}$ such that $s_n\nearrow f$ and using the Monotone Convergence Theorem.

If $f$ is a real valued function, we can apply (**) to $|f|$, which implies that $f$ is integrable with respect to $\mu T^{-1}$ if $f\circ T$ is integrable with respect to $\mu$. The result then follows by decomposing $f$ to its positive and negative parts and using the first part of the proposition on these two non-negative functions.