Riemann Integral-Questions

Hello! (Wave)

I am looking at the Riemann integral and I have two questions.

Theorem: Let $f: [a,b] \to \mathbb{R}$ bounded and $c \in (a,b)$. Then $f$ is integrable in $[a,b]$ iff it is integrable in $[a,c]$ and in $[c,b]$. In this case we have $\int_a^b f=\int_a^c f + \int_c^b f$.

At the proof, we use the Riemann criterion to conclude that $\mathcal{U}(f,P)-\mathcal{L}(f,P)<\epsilon$, so $f$ is integrable.

Then the following is stated:

We note that the quantities

$$\int_a^b f \text{ and } \int_a^c f + \int_c^b f$$

are between the numbers $\mathcal{L}(f,P)$ and $\mathcal{U}(f,P)$. Thus,

$$\left| \int_a^b f- \int_a^c f - \int_c^b f\right| \leq \mathcal{U}(f,P)-\mathcal{L}(f,P) < \epsilon.$$

Since the relation holds for any $\epsilon>0$, we get that $\int_a^b f=\int_a^c f+ \int_c^b f$.I haven't understood why the quantities$\int_a^b f \text{ and } \int_a^c f + \int_c^b f$ are between the numbers $\mathcal{L}(f,P)$ and $\mathcal{U}(f,P)$.
Could you explain it to me?

My second question is from the proof of the theorem that if $f,g: [a,b] \to \mathbb{R}$ integrable, then $f+g$ is integrable and $\int_a^b (f+g)=\int_a^b f +\int_a^b g$.

At the proof, we consider a partition of $[a,b]$, $\mathcal{P}=\{ a=t_0< \dots<t_n=b\}$.

Then $\mathcal{U}(f+g,P) \leq \mathcal{U}(f,\mathcal{P})+\mathcal{U}(g,\mathcal{P})$, $\mathcal{L}(f+g,P) \geq \mathcal{L}(f,\mathcal{P})+\mathcal{L}(g,\mathcal{P})$.

Now let $\epsilon>0$. SInce $f,g$ are integrable, there are partitions $\mathcal{P}_1, \mathcal{P}_2$ of $[a,b]$ such that

$$\mathcal{U}(f,\mathcal{P}_1)-\mathcal{L}(f,\mathcal{P}_1)< \frac{\epsilon}{2}, \mathcal{U}(g,\mathcal{P}_2)-\mathcal{L}(g,\mathcal{P}_2)< \frac{\epsilon}{2}$$

Then, if we set $\mathcal{P}=\mathcal{P}_1 \cup \mathcal{P}_2$, we have

$$\mathcal{U}(f,\mathcal{P})-\mathcal{L}(f,\mathcal{P})< \frac{\epsilon}{2}, \mathcal{U}(g,\mathcal{P})-\mathcal{L}(g,\mathcal{P})< \frac{\epsilon}{2}$$

From these relations we get that

$$\int_a^b f < \mathcal{L}(f, \mathcal{P})+\frac{\epsilon}{2}, \int_a^b g< \mathcal{L}(g, \mathcal{P})+\frac{\epsilon}{2}$$

and

$$\int_a^b f > \mathcal{U}(f, \mathcal{P})-\frac{\epsilon}{2}, \int_a^b g> \mathcal{U}(g, \mathcal{P})-\frac{\epsilon}{2}.$$

By adding the first two relations, we get that $\int_a^b f+\int_a^b g \leq \int_{\underline{a}}^{b} (f+g)+\epsilon$.

$\dots$Could you explain to me how we get that

$$\int_a^b f < \mathcal{L}(f, \mathcal{P})+\frac{\epsilon}{2}, \int_a^b g< \mathcal{L}(g, \mathcal{P})+\frac{\epsilon}{2}$$

and

$$\int_a^b f > \mathcal{U}(f, \mathcal{P})-\frac{\epsilon}{2}, \int_a^b g> \mathcal{U}(g, \mathcal{P})-\frac{\epsilon}{2}.$$

? :unsure:

evinda said:

I haven't understood why the quantities$\int_a^b f \text{ and } \int_a^c f + \int_c^b f$ are between the numbers $\mathcal{L}(f,P)$ and $\mathcal{U}(f,P)$.
Could you explain it to me?

Hey evinda!

Suppose we restrict the partition $P$ to the interval [a,c], which we'll call $P_1$.
Then $\mathcal{L}(f,P_1)\le\int_a^c f\le\mathcal{U}(f,P_1)$ isn't it? :unsure:

Suppose we do the same for the interval [c,b], can we find the inequality then? (Wondering)

Hi there! I'll try my best to explain the answers to your questions.

For your first question, we can use the Riemann criterion to show that the quantities $\int_a^b f$ and $\int_a^c f + \int_c^b f$ are between $\mathcal{L}(f,P)$ and $\mathcal{U}(f,P)$. This is because the Riemann criterion states that for a bounded function $f$ to be integrable on $[a,b]$, the upper and lower sums of $f$ for any partition $P$ must be arbitrarily close. In other words, for any $\epsilon>0$, we can find a partition $P$ such that $\mathcal{U}(f,P)-\mathcal{L}(f,P)<\epsilon$. This shows that the upper and lower sums are "sandwiched" between $\mathcal{L}(f,P)$ and $\mathcal{U}(f,P)$, and therefore the integrals $\int_a^b f$ and $\int_a^c f + \int_c^b f$ must also be between these numbers.

For your second question, we can use the fact that $f$ and $g$ are integrable to show that $\int_a^b f < \mathcal{L}(f, \mathcal{P})+\frac{\epsilon}{2}$ and $\int_a^b g< \mathcal{L}(g, \mathcal{P})+\frac{\epsilon}{2}$. This is because, by definition of integrability, we can find partitions $\mathcal{P}_1$ and $\mathcal{P}_2$ such that $\mathcal{U}(f,\mathcal{P}_1)-\mathcal{L}(f,\mathcal{P}_1)< \frac{\epsilon}{2}$ and $\mathcal{U}(g,\mathcal{P}_2)-\mathcal{L}(g,\mathcal{P}_2)< \frac{\epsilon}{2}$. Then, by setting $\mathcal{P}=\mathcal{P}_1 \cup \mathcal{P}_2$, we can show that $\mathcal{U}(f,\mathcal{P})-\mathcal{L}(f,\mathcal{P})< \frac{\epsilon}{2}$ and $\mathcal{U