Question

whats the difference between a function being piecewise continous and continous almost everywhere

Answer 1

I think they both mean that they are discointious at at most finittely many points

Answer 2

A piece-wise continuous function has some finite number of "breaks" in it and doesn’t diverge to infinity anywhere.

Continuous almost everywhere, informally, means that if you choose any (random) point the probability that it's continuous is exactly 1.

Answer 3

the second statement is kind of confusing

Answer 4

for example, what kind of category is heaviside function?

Answer 5

Heaviside is almost everywhere.

Answer 6

why is it and why not piecewise? what preveents heaviside from ebing piecewise?

Answer 7

Heaviside Step Function is Piecewise Continuous, it has finite breaks, and it is also continuous almost everywhere. (Sorry)

Answer 8

ok, I can see it being piecewise but not almost everywhere, I mean how does it fits the definitiotn of choosing a random point and get probablity 1 for continous. It is hard for me to apply this definition to heaviside

Answer 9

Choose any point, and the probability that the function is continuous at that point is exactly one, because you have 1 break, and an infinite number of values for which it's continuous.

Answer 10

probably Im thinking the wrong way but if I happen to choose x=0, isnt it discontious?

Answer 11

Yes, but you are choosing a particular point.
Out of all points, if you were to choose randomly, what is the chance you will choose a continuous point?

Answer 12

100% Oh thatmakes sense now!

Answer 13

:)

Answer 14

so if a fucntion is continous a.e., then that means it have at most contable infinitly many dicontiously points. T or F

Answer 15

I know for it to be piecewise, you cant have infinitly many discontiuity. so How about almost everywhere

Answer 16

think of a output made up of smaller sections the formula for the output depends on the set of input values.

Answer 17

**an output

Answer 18

This function is continuous almost everywhere, but it is not piece wise continuous, because it is discontinuous at infinite number of breaks.
\(\color{black}{\displaystyle f(x)=\begin{cases}2 & x \in \mathbb{Q}\\1& x \notin \mathbb{Q}\end{cases} }\)

Answer 19

(Because \(\color{black}{\displaystyle m\left(\mathbb{Q}\right)=0 }\))

Answer 20

you can do same for other sets with cardinality \(\aleph_0\) for example.

Answer 21

so at what point is a function no longer continous a.e, the measure of Q is 0 kind of confuses me, I only heard about the concept of measure and never quite understand it. I guess it does matter since I wont study any grad level analysis

Answer 22

*doesnt

Answer 23

\(\color{black}{\displaystyle f(x)=\begin{cases}2 & x \in \mathbb{R}\\1& x \notin \mathbb{R}\end{cases} }\)

Answer 24

\(\color{black}{\displaystyle m(\mathbb{R})\ne0 }\)
(it has to be countable, but \(\color{black}{\mathbb{R} }\) is not, as it is not denumerable.)

Answer 25

((and of course, it is not finite.))

Answer 26

if x is not part of R then it has to be C/R, I forgto which way the slash goes but you get the idea. I am thinking this becuase our input has to be "something" right, and if it is not real it has to be complex with reals excluded

Answer 27

Yes, (\color{black}{\mathbb{R}^c=\mathbb{C}-\mathbb{R}}\).

Answer 28

Yes, \(\color{black}{\mathbb{R}^c=\mathbb{C}-\mathbb{R}}\). oops:)

Answer 29

but then if we pick a random point, how does it come out with probability of being contiusous less than 1

Answer 30

basically Im trying to apply the defintion to this example and without getting into all the measure thingy, cause thats tooo complicated for me

Answer 31

The definition was bad. I attempted to distinguish between PW continuous and AE continuous. Suppose your function is not continuous over a set \(\color{black}{A }\). Then, if \(\color{black}{m(A)=0 }\) then it's continuous AE.

Answer 32

ok, guess we have to go to measure for it to work huh

Answer 33

Yes, Lebesgue-measure.

Answer 34

(Did I spell it right?)

Answer 35

Well, guess thats where I'll stop cause I am not doing more analysis than watching the undergraduate course from NCTU

Answer 36

one last question before I stop bothing you. how does the condition of diffitiable and integrateble connect with PW and AE contiously

Answer 37

so I am asking what sort of conditions must a PW contious fucntion have in order for it to be diffitiable? integrateble? How about AE functions?

Answer 38

sorry,I realized its really four questions

Answer 39

I want to say, it is the same as differentiable and integrable as for regular functions, except that you are talking about whether or not you can integrate and differentiate it over each of its breaks.

Answer 40

Also as a note: If your interval is [a,b], it will be differentiable over (a,b) (and not at endpoints).

Answer 41

\(f(x)=\left|x+1\right|\) is differentiable almost everywhere as an example.

Answer 42

thank you so mcuh for the answers today! I'll tag you next time if I have some other concepts I dont understand

Answer 43

Will do my best; I don't know any math, frankly.

Answer 44

yw