Show that $f$ is integrable using Archimedes-Riemann theorem

Question

Define $$f(x)=\begin{cases} x, 2 \leq x \leq 3 \\ 2, 3 < x \leq 4 \end{cases}$$ Show that $f$ is integrable on $[2,4]$ using Archimedes-Riemann theorem (Hint: find an Archimedian sequence of partitions that all have 3 as a partition point).

My attempt: I tried to use the regular partition, but it skips the point 3, so I'm thinking of a partition with $x_i-x_{i-1}=\frac{1}{n}.$ Then $m_i=\inf\{f(x)|x in [x_{i-1},x_i]\}=x_{i-1}, i =1,..., n, m_i = 2, i = n+1,...,2n.$ $M_i=\sup\{f(x)|x in [x_{i-1},x_i]\}=x_{i}, i =1,..., n, M_i = 2, i = n+1,...,2n.$ Therefore, $$L(f,P_n)=\sum_{i=1}^{2n}m_i(x_i-x_{i-1})=2*\frac{1}{n}+(2+\frac{1}{n})*\frac{1}{n}+...+(2+\frac{n-1}{n})*\frac{1}{n}+2*\frac{1}{n}+...+2*\frac{1}{n}=\frac{9n-1}{2n}.$$ Similarly, $U(f,P_n)=\sum_{i=1}^{2n}M_i(x_i-x_{i-1})=\frac{9n+3}{2n}.$

$\lim_{n \rightarrow \infty} L(f,P_n)=\lim_{n \rightarrow \infty} U(f,P_n)=\frac{9}{2},$ so $f$ is integrable.

Is it correct? I noticed that what I've done actually is showing that $x$ is integrable on subinterval $[2,3]$ and 2 is integrable on $[3,4]$, so I'm kind of confused and I suppose this is not the right approach. Maybe I should've chosen another sequence of partitions... Correct me if I'm wrong.

Answer 1

I noticed that what I've done actually is showing that $x$ is integrable on subinterval $[2,3]$ and $2$ is integrable on $[3,4]$, so I'm kind of confused and I suppose this is not the right approach.

Well, actually, that's exactly what's going on. That's the function we have.
Theorem: Given $a<b<c$, a function $f$ is (Riemann) integrable on $[a,c]$ if and only if it is integrable on both $[a,b]$ and $[b,c]$, and $\int_a^c f = \int_a^b f+\int_b^c f$.

And when we have a piecewise defined function, with multiple formulas? The standard practice to evaluate it is to carve up the interval so as to deal with one formula at a time. Is it any surprise we can do this even when we go down to the definitions?

One quibble: $M_{n+1}=\sup\{f(x)|x\in [x_n,x_{n+1}]\}$ is equal to $3$, not $2$. After all, the lower endpoint of that interval is $3$, and $f(3)=3$. Fortunately, that only applies to a single subinterval, with a width that goes to zero.