A step function integrated by additivity over subintervals, and the same value from the definition

Let $f : [0,2] \to \mathbb{R}$ be

$f(x) \;:=\; \begin{cases} 1 & 0 \le x < 1, \\ 3 & 1 \le x \le 2. \end{cases}$

Then $f$ is integrable on $[0,2]$ and

$\int_0^2 f \;=\; 1 + 3 \;=\; 4 .$

Two routes are carried out below. The first splits at the jump with For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then $\int_a^b f = \int_a^c f + \int_c^b f$; with the oriented form for arbitrary $a,b,c$ and evaluates each piece with Changing an integrable function at finitely many points changes neither its integrability nor its integral and the constant case of If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$. The second computes lower and upper Darboux sums directly on a three-point partition and squeezes.

Why this is the smallest interesting example. The jump sits at the interior point $1$, which is a partition point of both routes, and a partition point lies in two subintervals (Partition of $[a,b]$ as a finite strictly increasing list $a = t_0 < t_1 < \dots < t_n = b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions). That is exactly the case singled out in the proof of Changing an integrable function at finitely many points changes neither its integrability nor its integral, and it is why the value of $f$ at the single point $1$ is irrelevant to both the integrability and the value while the upper sums still see it.

• The two sums of $P_{\eta}$ are not symmetric, and that is a statement about these partitions. For each $\eta \in (0,1)$ the lower sum $L(f,P_{\eta})$ is exactly $4$ while the upper sum is $4 + 2\eta > 4$, so this family already locates the integral exactly from below while its upper sums only approach it. Nothing is claimed here about an arbitrary partition: a partition whose points do not include $1$ has a subinterval on which $f$ takes both values, and its lower sum is then strictly below $4$.

• The value at the jump is irrelevant to the integral, but not to the sums. Replacing $f(1) = 3$ by any real leaves $f$ integrable with the same integral, by Changing an integrable function at finitely many points changes neither its integrability nor its integral. What it changes is which partitions realise the extremes: with $f(1) := 1$ the subinterval $[1-\eta,1]$ carries $m_1 = M_1 = 1$, while $[1,2]$ now carries $m_2 = 1$ and $M_2 = 3$, so $L(f,P_{\eta}) = 2$ and $U(f,P_{\eta}) = 4$ and the asymmetry moves to the other side. The integral is $4$ either way.

• Additivity is what makes the first route short. Without For $a<c<b$: $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$, and then $\int_a^b f = \int_a^c f + \int_c^b f$; with the oriented form for arbitrary $a,b,c$ the only route is the second one, which needs the Darboux sums of a function that is not constant on any partition containing the jump in the interior of a subinterval.