Sufficiency of Lebesgue's Criterion for Riemann Integrability

Question

In my notes there is a proof of the the following:

Proposition. A function $f:[a,b]\to \mathbb{R}$ is Riemann integrable on $[a,b]$ if $f$ is bounded and

$$\text{disc}(f):=\{x\in [a,b] : f \text{ is discontinuous at } x \}$$

has measure zero.

However, am having trouble with some parts of the proof which I have highlighted in bold:

Proof. Suppose $f$ is bounded and $\text{disc}(f)$ has measure zero. Let $M=\sup_{[a,b]}|f|$ and let $\epsilon>0$ be given. By definition of measure zero, there exist $(a_j,b_j)$, $j\geq 1$, such that $\text{disc}(f)\subset \bigcup_{j\geq 1} (a_j,b_j)$ and $\sum_{j=1}^{\infty}(b_j-a_j)<\epsilon$. Define

$$A:=\bigcup_{j\geq 1} (a_j,b_j) \hspace{1cm} K:=[a,b]\setminus A$$

Note that $K$ is closed and bounded, hence compact. Also, continuity implies that, for each $x\in K$, there exist $\delta(x)=\boldsymbol{\delta(x,\epsilon)\in (0,\epsilon)}$ such that

$$ y\in[a,b], |x-y|\leq\delta(x) \implies |f(x)-f(y)|\leq \epsilon \hspace{0.5cm}(*)$$

The collection of open intervals $(x-\delta(x),x+\delta(x))$, $x\in K$, covers $K$. Since $K$ is compact, there exist finitely many elements of $K$, say $k_1,\dots,k_r\in K$, such that

$$K\subset \bigcup_{s=1}^r (k_s-\delta_s,k_s+\delta_s)$$ where $\delta_s:=\delta(k_s)$. Let $L$ be the set of all points of the form $k_s\pm \delta_s$ and consider the partition

$$P:=\{a,b\}\cup\{x\in L:a\leq x\leq b\}$$

of $[a,b]$. Let $\mathcal{I}(P)=\{I_1,\dots,I_n\}$ be the compact intervals determined by this partition. For $I_j\in \mathcal{I}(P)$ we distinguish between two cases:

Case 1. There exist $1\leq s\leq r$ such that $I_j\subset[k_s-\delta_s,k_s+\delta_s]$. We call $\mathcal{J}_{good}$ the set all such $j$. For $j\in\mathcal{J}_{good}$ we have, using $(*)$ above,

$$|f(x)-f(y)|\leq |f(x)-f(k_s)|+|f(k_s)-f(y)|\leq \epsilon+\epsilon=2\epsilon$$

for all $x,y\in I_j$. Hence $\text{osc}_f (I_j):=\sup_{x,y\in I_j}|f(x)-f(y)|\leq 2\epsilon$.

Case 2. There is no $1\leq s\leq r$ such that $I_j\subset[k_s-\delta_s,k_s+\delta_s]$. We call $\mathcal{J}_{bad}$ the set all such $j$. For $j\in\mathcal{J}_{bad}$ we have $\text{osc}_f (I_j)\leq 2M$. We claim that

$$\sum_{j\in\mathcal{J}_{bad}} |I_j|\leq 3\epsilon$$

Let $I_j=[x_{j-1},x_j]$. If $x_{j-1}=a$ or $x_{j}=b$ we have the bound $\boldsymbol{|I_j|\leq\epsilon}$. Hence we only have to show that

$$\sum\limits_{\substack{j\in\mathcal{J}_{bad} \\ a,b\notin I_j}} |I_j|\leq \epsilon \hspace{0.5cm} (**)$$

We know that in this case $x_{j-1},x_j \in L$. But then the only way that we can have $I_j\not\subset[k_s-\delta_s,k_s+\delta_s]$ for some $s$ is to have

$$I_j\subset [a,b]\setminus \bigcup_{s=1}^r (k_s-\delta_s,k_s+\delta_s)$$

and since $K\subset \bigcup_{s=1}^r (k_s-\delta_s,k_s+\delta_s)$ we find that

$$I_j\subset[a,b]\setminus K\subset A=\bigcup_{j\geq 1} (a_j,b_j) $$

Hence $$\bigcup\limits_{\substack{j\in\mathcal{J}_{bad} \\ a,b\notin I_j}}\subset \bigcup_{j\geq 1} (a_j,b_j) $$

Because the intervals $I_j$ have disjoints interiors, $(**)$ now follows from $\sum_{j=1}^{\infty}(b_j-a_j)<\epsilon$.

The proof then concludes using Darboux's criterion.

As you can see I have three issues:

1. Why do we need $\delta(x)<\epsilon$?
2. Why is the bound $|I_j|\leq \epsilon$ valid if $j\in\mathcal{J}_{bad}$ and $x_{j-1}=a$ or $x_j=b$?
3. Why does $\sum\limits_{\substack{j\in\mathcal{J}_{bad} \\ a,b\notin I_j}} |I_j|\leq \epsilon $ follows from $\sum_{j=1}^{\infty}(b_j-a_j)<\epsilon$?

Any help is greatly appreciated.

Answer 1

1. I think this is a trick that is used in this proof ( just like $\epsilon 2^{-n}$ trick used in a lot of measure-theoretic results).

2. Suppose $x_{j-1} =a$. Clearly $a \notin L$ so $x_j$ must be either $k_s+\delta_s$ or $k_s-\delta_s $ for some $1\leq s\leq r$ (Because the endpoints of $I_js$ are formed by the points in $P$). In either case, there exists $k_{s'} \in K$ such that $|I_j| \leq |(a+\delta_{s'})-a|$, where $s'\leq r$. By the choice of $\delta_{s'} (=\delta(k_{s'}))\,$, $|I_j| \leq \epsilon$.

3. $I_js $ have disjoint interiors. So the result follows from the countable-additivity, sub-additivity & the monotonicity of the Lebesgue measure.

$|\cup I_j| = \sum |I_j| \leq |\cup (a_j , b_j)| \leq \sum (b_j - a_j) < \epsilon$ (the last inequlity is due to the choice of $(a_j ,b_j)$s)

Answer 2

This is a result by Lebesgue:

Theorem: A function $f$ is Riemann--integrable in $[a,b]$ iff $f$ is bounded and continuous $\lambda$--a.s. in $[a,b]$.

Necessity is proved here.

The proof of sufficiency relies on some technical results that analyze the set where the integrand $f$ is continuous.

I present an online of these technical results below

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Let $h$ be a function on an arbitrary space $X$ with values in a metric space $(S',d')$. For any $T\subset S$, the modulus of continuity of $h$ on $T$ is defined as $$\Omega_h(T):=\sup\{\rho'(h(x),h(y)):x,y\in T\}.$$ If $X$ is also a metric space, the modulus of continuity $h$ at $x$ is defined as $$\omega_h(x)=\lim_{\delta\searrow0}\Omega_h(B(x;\delta))=\inf_{\delta>0}\Omega_h(B(x;\delta))$$

Lemma 1: Let $S$ and $S'$ be metric spaces and let $h:S\rightarrow S'$. For any $r>0$, the set $J_r=\{x\in S:\omega_h(x)\geq r\}$ is closed.

Here is a short proof:

If $x\in J^c_r$, there is $\delta>0$ such that $\Omega_h(B(x;\delta))<r$. Clearly $B(x;\delta)\subset J^c_r$.

Lemma 2: For any function $h:S\longrightarrow S'$, the set $D_h\subset S$ of discontinuities of $h$ is a $\sigma$--F set and thus, Borel measurable.

Here is a short proof:

$h$ is continuous at $x$ if an only if $\omega_h(x)=0$. By Lemma 1 the set $J_\varepsilon= \{x\in S:\omega(x)\geq\varepsilon\}$ is closed in $S$. Therefore $D_h=\bigcup_n J_{1/n}$ is a $\sigma$--F set.

Lemma 3: If $\omega_f(x)<\varepsilon$ for all $x\in[c,d]\subset[a,b]$, then exists $\delta>0$ such that $\Omega_f(T)<\epsilon$ for all $T\subset[c,d]$ with $\operatorname{diam}(T)<\delta$.

Here is a short proof:

For any $x\in[c,d]$ there is $\delta_x>0$ such that $\Omega_f(B(x;\delta_x)\cap[c,d])<\varepsilon$. The collection of all $B(x;\delta_x/2)$ forms an open cover of $[c,d]$. By compactness, there are $x_1,\ldots,x_k$ with $[c,d]\subset\bigcup^k_{j=1}B(x_j;\delta_j/2)$. Let $\delta=\min\{\delta_j/2\}$. If $T\subset[c,d]$ with $\text{diam}(T)<\delta$, then is fully contained in at least one $B(x_j;\delta_j)$ so $\Omega_f(T)<\epsilon$.

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Here is a proof of sufficiency of Lebesgue's theorem:

For each $r>0$, define $J_r=\{x\in[a,b]:\omega_f(x)\geq r\}$. Each $J_r$ is a closed subset in $[a,b]$ (see Lemma 1) and the set of discontinuities of $f$ is given by $\mathcal{J}=\bigcup_{k\in\mathbb{N}}J_{1/k}$. Each $J_{1/k}$, being a compact subset of measure zero, is covered by the union $A_k$ of a finite collection of open intervals in $[a,b]$ whose lengths add up to something less than $\tfrac1k$. Clearly $B_k=[a,b]\setminus A_k$ is the union of a finite collection of close subintervals in $[a,b]$. By Lemma 3, there is $\delta_k>0$ such that if $T\subset[a,b]\setminus A_k$ and $\text{diam}(T)<\delta_k$, then $\Omega_f(T)<\tfrac1k$. Let $\mathcal{P}_k$ be a partition formed by the subintervals of length less than $\delta_k$, and whose endpoints are with in $\mathcal{A}_k$ to in $B_k$. It follows that $$ U(f,\mathcal{P}_k)-L(f,\mathcal{P}_k)=S_1+S_2 $$ where $S_1$ is formed by the subintervals containing points of $J_k$ and $S_2$ by subintervals contained in $B_k$. As $S_1\leq (M-m)/k$ and $ S_2\leq (b-a)/k$, for $k$ large enough we have that $U(f,\mathcal{P}_k)-L(f,\mathcal{P}_k)<\varepsilon$.