Necessary and Sufficient Conditions for Riemann Integrability

Solution 1:

Assertion $(a)$ is true (and continuity over a closed interval implies boundedness), but assertion $(b)$ is not! Take $${\bf 1}_{[0,1]\cap \Bbb Q}$$

It isn't Riemann integrable over $[0,1]$, yet it is bounded.

ADD The conditions for Riemann integrability are very precise. A (bounded) function is Riemann integrable over a closed interval $[a,b]$ if the following equivalent conditions hold:

$(1)$ For each $\epsilon>0$ there exist step functions $s_1\leq f \leq s_2$ such that $$\int_a^b s_2-\int_a^b s_1<\epsilon$$

$(2)$ There exists a number $I$ (the integral) such that for each $\epsilon >0$ there exists a $\delta >0$ such that for each tagged partition $P=\{x_0,\dots,x_n,t_0,\dots,t_n\}$ of $[a,b]$ with $\Delta P<\delta $ (the mesh of $P$) we have $$\left| I-\sum_{x,t\in P}f(t)\Delta x\right| <\epsilon$$

$(3)$ For each $\epsilon >0$ there exists a partition $P_\epsilon=\{x_0,\dots,x_n\}$ of $[a,b]$ such that $$U(f,P_\epsilon)-L(f,P_\epsilon)<\epsilon$$

$(4)$ It holds that $$\sup\{L(f,P):P \text{ is a partition of } [a,b]\}=\inf\{U(f,P):P \text{ is a partition of } [a,b]\}$$

$(5)$ The set of discontinuities of $f$ has Lebesgue measure $0$, that is, given $\epsilon >0$, the set $$A=\{x\in[a,b]:f\text{ is discontinuous at } x\}$$ can be covered my countably many open intervals such that the sum of their lengths is less than $\epsilon$.

Solution 2:

Note that, the statement in $(b)$ is not true in general. To see this, just consider the Dirichlet function which is not Riemann integrable.

Solution 3:

Riemann integrability depends on $f$ being defined on a closed interval; so I'll assume that the domain is as given.

For question a, you want to consider the proposition that a continuous function on a compact set (like say a closed interval) is bounded. If you can show that, then it follows that the function is bounded and continuous AE on the domain.

For question b, I believe the statement is false, consider the characteristic function on the rationals over $[0,1]$. The upper sums are always 1 and the lower sums are always 0 no matter how you partition the domain.