Understanding the proof if the given function is Riemann integrable?

Question

I am trying to understand the proof if the function given in the picture below is integrable by using the Epsilon-Delta definition of Reimann integral. I was supposed to take real analysis course this semester but didn't. So I am self studying and I was going through Reimann integrals. As far as i understand, for any tagged partition P of [0,3], i split it into two disjoint partitions P_1 and P_2 where P_1 forms a partition of [0,1] having its tag within and similarly for P_2 having tags only in (1,3]. Now i understand that the subintervals in P_1 not necessarily are the same subintervals formed in P. But in the proof it seems like the subintervals are the same. By saying P_1 has its tags in [0,1], does it mean its a partition of [0,1] and not [0,3]? If someone can clear my misunderstandings or provide me with an explanation of this proof, I'd be grateful.

Answer 1

IF you took arbitary other partitions $\dot{\mathcal P}_1, \dot{\mathcal P}_2$, then you would be right that the subintervals and tags may not be the same as in $\dot{\mathcal P}$.

But that is not what they are doing. It is not their purpose here to investigate the integrals on $[0,1]$ or $[1,3]$, so they do not need arbitrary partitions. The point of $\dot{\mathcal P}_1$ and $\dot{\mathcal P}_2$ is to produce an estimate of $S(g;\dot{\mathcal P})$. They define $\dot{\mathcal P}_1$ to be the subintervals of $\dot{\mathcal P}$ with tags in $[0,1]$, and $\dot{\mathcal P}_2$ to be the remaining subintervals of $\dot{\mathcal P}$. The result is $\dot{\mathcal P} = \dot{\mathcal P}_1 \cup \dot{\mathcal P}_2$. The partition intervals are those of $\dot{\mathcal P}$ by definition.

Answer 2

You should try to understand the argument given by author without going into too much symbolism.

We are given a partition $\dot{\mathcal P} $ of $[0,3]$ with norm less than $\delta$. This means that we are given a set of sub-intervals $$[x_0,x_1],[x_1,x_2],\dots,[x_{n-1},x_n]$$ where $$0=x_0<x_1<x_2<\dots<x_n=3$$ We are also given some tags $t_i\in[x_{i-1},x_i]$ for $i=1,2,\dots,n$. And further for each $i$ we have $x_i-x_{i-1}<\delta$.

This also implies that $$t_1\leq t_2\leq \dots\leq t_n$$ The key step in the argument is to separate the tags $t_i$ which lie in $[0,1]$ from those which lie in $(1,3]$. To make the argument explicit there exists a unique integer $k$ with $1\leq k\leq n$ such that $t_1,t_2,\dots,t_k$ all lie in $[0,1]$ and $t_{k+1},t_{k+2},\dots,t_n$ lie in $(1,3]$ (this part may be non existent if $k=n$).

Now all the sub-intervals $$[x_0,x_1],[x_1,x_2],\dots, [x_{k-1},x_k]$$ make up $\dot{\mathcal P} _1$ and $$[x_k, x_{k+1}],[x_{k+1},x_{k+2}],\dots,[x_{n-1},x_n]$$ make up $\dot {\mathcal P} _2$ (which may be empty as remarked earlier).

Now we can note that $$x_k - t_k\leq x_k-x_{k-1}<\delta$$ so that $$x_k<t_k+\delta\leq 1+\delta$$ Further note that $$t_{k+1}-x_{k}\leq x_{k+1}-x_k<\delta$$ so that $$x_k>t_{k+1}-\delta>1-\delta$$ It thus follows that the union of sub-intervals in $\dot {\mathcal P} _1$ contains $[0,1-\delta]$ properly and is properly contained in $[0,1+\delta]$.

A similar analysis can be done for $\dot{\mathcal P} _2$.