Will ##M_i = m_i## if an interval is made vanishingly small?

[Adesh]

TL;DR

If I have an interval $[x_{i-1}, x_i]$ and if it's length $x_i - x_{i-1}$ is made as small as we desire then will we have $M_i = m_i$?

We define :
$$M_i = sup \{f(x) : x \in [x_{i-1}, x_i ] \}$$
$$m_i = inf \{f(x) : x \in [ x_{i-1}, x_i ] \}$$
Now, if we make the length of the interval $[x_{i-1}, x_i]$ vanishingly small, then would we have $M_i = m_i$? I have reasons for believing so because as the size of the interval is decreasing we will be having less choice for $inf$ and $sup$ and therefore $inf$ will increase and $sup$ will decrease, so can we made them arbitrarily close by taking a sufficiently small lengthed interval?

Thank You.

 

[PeroK]

Now, if we make the length of the interval $[x_{i-1}, x_i]$ vanishingly small,

There is no such thing in real analysis. Heuristically, in physics we can take things to be "vanishingly small", but not in the rigorous mathematics of $sup$ and $inf$.

 

[Adesh]

There is no such thing in real analysis. Heuristically, in physics we can take things to be "vanishingly small", but not in the rigorous mathematics of $sup$ and $inf$.

Can’t we make $gap~P$ as small as we want by adding more and more points?

 

[PeroK]

Can’t we make $gap~P$ as small as we want by adding more and more points?

Yes, but it's still always a finite interval.

 

[Adesh]

Yes, but it's still always a finite interval.

Will that affect the difference $M_i - m_i$?

 

[PeroK]

Will that affect the difference $M_i - m_i$?

Affect it in what way? You have $sup \ f$ and $inf \ f$ on a finite interval.

 

[Adesh]

Affect it in what way? You have $sup \ f$ and $inf \ f$ on a finite interval.

My thinking says that as intevral's length gets shorter and shorter the difference $M_i - m_i$ will get smaller and smaller. Is that true?

 

[PeroK]

My thinking says that as intevral's length gets shorter and shorter the difference $M_i - m_i$ will get smaller and smaller. Is that true?

Not necessarily. If one interval is a subset of another then the difference between sup and inf cannot be greater for the subset. But, the difference between sup and inf might stay the same.

 

[Adesh]

If one interval is a subset of another then the difference between sup and inf cannot be greater for the subset.

Won't the difference going to decrease for the subset? It may stay the same or it may decrease because for a very very small interval our function will have very few choice for inf and sup.

 

[PeroK]

Won't the difference going to decrease for the subset? It may stay the same or it may decrease because for a very very small interval our function will have very few choice for inf and sup.

Any open interval, no matter how small, is analytically equivalent to the set of all real numbers. There's really no such thing as a "small" interval. You need to rethink your approach to this mathematics.

 

[Stephen Tashi]

My thinking says that as intevral's length gets shorter and shorter the difference $M_i - m_i$ will get smaller and smaller. Is that true?

"gets shorter and shorter"?

Your are thinking about a process, presumably taking place in time and in a sequence of steps. You need to look at $L_M = lim_{i \rightarrow \infty} M_i$ Trying to talk about $M_i$ as if it is one specific number defined for one specific index and then saying it also changes is a contradiction.

 

[pasmith]

Summary:: If I have an interval [xi−1,xi] and if it's length xi−xi−1 is made as small as we desire then will we have Mi=mi?

We define :
Mi=sup{f(x):x∈[xi−1,xi]}
mi=inf{f(x):x∈[xi−1,xi]}
Now, if we make the length of the interval [xi−1,xi] vanishingly small, then would we have Mi=mi? I have reasons for believing so because as the size of the interval is decreasing we will be having less choice for inf and sup and therefore inf will increase and sup will decrease, so can we made them arbitrarily close by taking a sufficiently small lengthed interval?

Thank You.

If f is continuous then for every \epsilon > 0 there exists a \delta > 0 such that if x_i - x_{i-1} < \delta then M_i - m_i < \epsilon. This is a simple consequence of the definition and the fact that a function which is continuous on a closed bounded interval attains its bounds.

If \{A_n\} is a sequence of nested subsets such that A_{n+1} \subsetneq A_n then f(A_{n+1}) \subset f(A_n). Hence \sup f(A_{n+1}) \leq \sup f(A_n) as removing points from a set might decrease its upper bound but cannot possibly increase it. Similarly \inf f(A_{n+1}) \geq \inf f(A_n). However the limits don't have to be equal: consider <br /> f : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases}<br /> -1 &amp; x &lt; 0 \\<br /> 0 &amp; x = 0 \\<br /> 1 &amp; x &gt; 0 \end{cases} with A_n = [-\frac1n, \frac1n]. Here \inf f(A_n) = -1 and \sup f(A_n) = 1 for every n.

 

[Adesh]

I also found (from someone’s help) signum function is counter example to my claim. Let $f$ be a signum function then any interval containing zero, no matter how small will $M_i -m_i =2$.

 

[FactChecker]

The signum function is just the tip of the iceberg. It can be solved with some restrictions on the choice of the $x_i$. But there are more difficult examples. Consider the function which is 0 for rational numbers and 1 for irrational numbers. This will eventually lead to an entirely new definition of integration (Lebesque integration) that is applicable to a much larger set of functions.

You should not worry about Lebesque integration now. It is mostly for theoretical use. For now, consider how things work with continuous or piecewise continuous functions.

 

[Adesh]

The signum function is just the tip of the iceberg. It can be solved with some restrictions on the choice of the $x_i$. But there are more difficult examples. Consider the function which is 0 for rational numbers and 1 for irrational numbers. This will eventually lead to an entirely new definition of integration (Lebesque integration) that is applicable to a much larger set of functions.

You should not worry about Lebesque integration now. It is mostly for theoretical use. For now, consider how things work with continuous or piecewise continuous functions.

Okay, I shall work with continuous or piecewise continuous functions for now. Thank you.