Prove that $\int f\ d\lambda = \int_{a}^{b} f(x)\ dx,$ for any $f \in \mathcal R[a,b].$

Theorem $:$ Let $f : [a,b] \longrightarrow \Bbb R$ be a Riemann integrable function. Then $f \in L_1[a,b]$ and $$\int f\ d\lambda = \int_{a}^{b} f(x)\ dx.$$

The proof given in my book is as follows $:$

At the fag end of the proof $f$ has been proved to be measurable which uses the completeness of the measure space $\left ([a,b], \mathcal L \cap [a,b], \lambda |_{[a,b]} \right ).$ Then it has been stated that $f \leq \Phi_n,$ $\forall$ $n$ and hence $f \in L_1[a,b]$ since $\Phi_n \in L_1[a,b]$ by the proposition $5.4.3 \ $(i) which states the following $:$

Let $f \in \Bbb L$ and $g \in L_1[a,b].$ If $|f(x)| \leq g(x)$ for a.e. $x(\mu).$ Then $f \in L_1[a,b].$

I don't know how to use this proposition to claim that $f \in L_1[a,b].$ Also I don't understand the way dominated convergence theorem used here. What I know about dominated convergence theorem is as follows $:$

Let $(X, \mathcal S, \mu)$ be a complete measure space. Let $g \in L_1(X)$ and $\{f_n \}_{n \geq 1}$ be a sequence in $\Bbb L$ such that $|f_n(x)| \leq g(x)$ a.e. $x(\mu).$ Let $f_n(x) \to f(x)$ as $n \to \infty$ a.e. $x (\mu).$ Then $f \in L_1(X)$ and $$\int f\ d\mu = \lim\limits_{n \to \infty} \int f_n\ d\mu.$$

From here how to conclude that $$\Psi_n\ \bigg\uparrow\ f \implies \int f\ d\lambda = \lim\limits_{n \to \infty} \int \Psi_n\ d\lambda.$$

Please help me in this regard. Thanks in advance.

Solution 1:

Here is a solution along the lines of your text book. The key ingredient is dominated convergence. The lower and upper Riemann sums provides sequences of step functions that converge to the ingtegrand almost surely.

Consider the measure space $([a,b],\mathscr{B}([a,b]),\lambda)$. A partition of $[a,b]$ is finite set $P=\{a=t_0<\ldots<t_n=b\}$. Define $m_k= \inf\{f(t):t\in[t_{k-1},t_k]\}$ and $M_k=\sup\{f(t):t\in[t_{k-1},t_k]\}$. The lower and upper sums are defined by $$ \begin{align} L(f,P)&=&\sum^n_{k=1}m_k(t_k-t_{k-1})\tag{1}\label{lower-darboux}\\ U(f,P)&=&\sum^n_{k=1}M_k(t_k-t_{k-1})\tag{2}\label{upper-darboux} \end{align} $$ Let $\mathcal{P}$ the collection of all partitions of $[a,b]$.

I will use this definition for Riemann integrability

Definition:

A function $f:[a,b]\rightarrow\mathbb{R}$ is Riemann integrable if $$ \begin{align} \sup_{P\in\mathcal{P}}L(f,P)=\inf_{P\in \mathcal{P}}U(f,P) \tag{3}\label{darboux-int} \end{align} $$ The common value $A(f)$ in~\eqref{darboux-int} is called the Riemann integral of $f$ over $[a,b]$.

It is easy to see that for any partitions $P_1$ and $P_2$ of $[a,b]$ $$ L(f,P_1)\leq L(f,P_1\cup P_1) \leq U(f,P_1\cup P_2)\leq U(f,P_2) $$

It follows that $f$ is Riemann integrable over $[a,b]$ if and only if $f$ is bounded and for any $\varepsilon>0$ there is a partition $P_\varepsilon$ such that $$ \begin{align} U(f,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon\tag{4}\label{darboux2} \end{align} $$

Theorem: Suppose that $f$ is Riemann--integrable in $[a,b]$, and let $\mathscr{M}([a,b])$ be the Lebesgue $\sigma$--algebra. Then, $f\in L_1([a,b],\mathscr{M}([a,b]),\lambda)$ and $f$ is continuous $\lambda$--a.s. Moreover, $A(f)=\int_{[a,b]}f\,d\lambda$.

Here is a short proof

Choose partitions $\mathcal{P}_n\subset\mathcal{P}_{n+1}$ such that $$U(f,\mathcal{P}_n)-L(f,\mathcal{P}_n)<1/n\tag{0}\label{zero}$$ For each partition $\mathcal{P}_n$, let $m_{n,k}=\inf\{f(t):t\in[t_{n,k-1},t_{n,k}]\}$ and $M_{n,k}=\sup\{f(t):t\in[t_{n,k-1},t_{n,k}]\}$. Let $g_n$ and $h_n$ be defined by $g_n(a)=h_n(a)$; and $g_n(t)=m_{n,k}$, $h_n(t)=M_{n,k}$ on $t\in(t_{n,k-1},t_{n,k}]$. Clearly, $g_n\leq g_{n+1}\leq f\leq h_{n+1}\leq h_n$ on $[a,b]$, and $\int_{[a,b]}g_n=L(f,\mathcal{P}_n)\leq U(f,\mathcal{P}_n)=\int_{[a,b]}h_n$.

Dominated convergence and $\eqref{zero}$ implies $\int_{[a,b]}g(x)dx=\int_{[a,b]}h(x)dx=A(f)$; since $g=\lim_ng_n\leq f\leq \lim_nh_n=h$,
$$g=f=h\qquad\text{a.s.}$$ and so $f$ is measurable. Let $\mathcal{D}=\{t\in[a,b]:g(t)<f(t)\}$. Then, $f$ is continuous at every point $x\notin\bigcup_n\mathcal{P}_n\cup \mathcal{D}$.

Example

The function $f=\mathbb{1}_{[0,1]\setminus\mathbb{Q}}\in L_1([0,1])$ and $\int_{[0,1]}f\,d\lambda=1$; however, $f$ is not Riemann integrable in $[0,1]$ since $U(f,\mathcal{P})-L(f,\mathcal{P})=1$ for any partition $\mathcal{P}$ of $[0,1]$.

Example

Let $f=\mathbb{1}_{[0,1/2)} + 10^{10}\mathbb{1}_{\{1/2\}} + 3\,\mathbb{1}_{(1/2,1]}$. Using the dyadic partition $\mathcal{P}_n=\{\frac{k}{2^n}: k=0,\ldots,2^n\}$ to construct $g_n$ and $h_n$ as in the proof of the Theorem, one gets that $g_n$ converges to $f$ everywhere but $\{1/2\}$; $h_n$ does converge to $f$ everywhere; $\{1/2\}$ is the of discontinuity in this case.