Why is the Infimum of f in Any Subinterval Equal to 0?

[evinda]

Hi! :)
I am looking at the following exercise:
Let $f:[a,b] \to \mathbb{R}$ integrable at $[a,b]$,such that $f(r)=0$,for each rational number $r \in [a,b]$.Prove that $\int_a^b f(x) dx=0$.
We suppose the partition $P=\{ a=t_0<t_1<...<t_n=b\}$ of $[a,b]$
$\underline{\int_{a}^{b}} f(x)dx=sup \{ L(f,P): P $ $\text{ partition of } [a,b]\}$
$L(f,P)=\Sigma_{k=0}^{n-1}(t_{k+1}-t_{k}) \cdot inf f([t_k,t_{k+1}])$.
Why is $ inf f([t_k,t_{k+1}])$ equal to $0$?Is it because of the fact that,if,for example it was $-1$ ,the lower integral couldn't be equal to the upper integral,since $f(r)=0>-1$??

 

[Opalg]

Re: inf f=0,why?

Hi! :)
I am looking at the following exercise:
Let $f:[a,b] \to \mathbb{R}$ integrable at $[a,b]$,such that $f(r)=0$,for each rational number $r \in [a,b]$.Prove that $\int_a^b f(x) dx=0$.
We suppose the partition $P=\{ a=t_0<t_1<...<t_n=b\}$ of $[a,b]$
$\underline{\int_{a}^{b}} f(x)dx=sup \{ L(f,P): P $ $\text{ partition of } [a,b]\}$
$L(f,P)=\Sigma_{k=0}^{n-1}(t_{k+1}-t_{k}) \cdot inf f([t_k,t_{k+1}])$.
Why is $ inf f([t_k,t_{k+1}])$ equal to $0$?Is it because of the fact that,if,for example it was $-1$ ,the lower integral couldn't be equal to the upper integral,since $f(r)=0>-1$??

$ \inf f([t_k,t_{k+1}])$ need not be equal to $0$. But what you can say is that $0$ must lie between $ \inf f([t_k,t_{k+1}])$ and $ \sup f([t_k,t_{k+1}])$. Use that to show that $0$ must lie between the lower and upper sums of any partition.

 

[ThePerfectHacker]

Re: inf f=0,why?

There is another way of defining what it means to be integrable that avoids the use of supremums and infimums. This was Riemann's original definition which turns out being equivalent to the definition you do in advanced calculus.

Definition: Let $f$ be a real-valued function (of a real variable) defined on the interval $[a,b]$. We say that $f$ is integrable on $[a,b]$ if there exists a number $I$ such that for every $\varepsilon >$ there is a $\delta > 0$ such that for any partition $P = \{ x_0,x_1,...,x_n\}$ of $[a,b]$ with $|| P || < \delta$ we have:
$$ \left| \sum_{k=1}^n f(t_k) (x_k - x_{k-1}) - I \right| < \varepsilon $$
Where $t_k \in [x_{k-1},x_k]$ and $||P||$ is defined by $|| P || = \max_k (x_k - x_{k-1})$.
This number $I$, if it exists, it must be unique, and is called the "integral" of $f$ on $[a,b]$.

What I like about this definition is that it is shorter than the one you are used to. Now using this definition solving your problem is not hard. We want to show that $I = 0$ where $f$ has the property that $f(r) = 0$ for every rational $r\in [a,b]$. So pick any $\varepsilon > 0$, we know there is a $\delta > 0$ such that for all $P$ with $||P||$ the inequality above holds. Since the rational numbers are dense we can pick $r\in [x_{k-1},x_k]$ so that $f(r) = 0$. Thus, the summation has value equal to $0$, it follows from here that $|I| < \varepsilon$ for every $\varepsilon > 0$. Hence, $I = 0$.

 

[evinda]

Re: inf f=0,why?

$ \inf f([t_k,t_{k+1}])$ need not be equal to $0$. But what you can say is that $0$ must lie between $ \inf f([t_k,t_{k+1}])$ and $ \sup f([t_k,t_{k+1}])$. Use that to show that $0$ must lie between the lower and upper sums of any partition.

Why must $0$ lie between $ \inf f([t_k,t_{k+1}])$ and $ \sup f([t_k,t_{k+1}])$ ?I haven't understood it yet... (Wondering)(Thinking)

 

[blue_raver22]

Hi there! It's great to see that you're working on this exercise and thinking about the concepts of integral partitions and infimum. You are on the right track with your reasoning.

To answer your question, the reason why $inf f([t_k,t_{k+1}])$ is equal to $0$ is because of the given condition that $f(r)=0$ for all rational numbers $r \in [a,b]$. This means that for any interval $[t_k,t_{k+1}]$, the infimum of $f$ over that interval must be $0$, since there exists at least one rational number within that interval where $f$ is equal to $0$.

As you mentioned, if the infimum were anything other than $0$, then the lower integral would not equal the upper integral, which goes against the definition of the integral. This is because the integral represents the area under the curve, and since $f(r)=0$ for all $r \in [a,b]$, the area under the curve must also be equal to $0$.

I hope this helps clarify things for you. Keep up the good work!