MHB How Do Riemann Integrals Handle Function Splits and Summations?

[evinda]

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:

 

[I like Serena]

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)

 

[soloenergy]

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