# Function Bounded and Continuous on (0,1) but not Integrable!

#### evry1cndie

Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!

#### Drexel28

MHF Hall of Honor
Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!
What kind of integral is defined on an open interval?

evry1cndie

#### Jose27

MHF Hall of Honor
Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!
This isn't true for the Riemann integral, since we can extend it to a function in [0,1] as 0 in the endpoints, and it would only be discontinous there.

evry1cndie

#### mabruka

I dont think it is true either, if f is continuous and bounded over (0,1) then

$$\displaystyle \bigg|\int_0^1 f(x)dx\bigg|\leq\bigg|\int_{(0,1)} fd\lambda\bigg|\leq M \lambda((0,1))<\infty$$

so f would be integrable.

Thus it is important to remove some of the hypothesis:

continuity or boundedness or measure finite set

so we can have some functions satisfying the statement that is left after the removal.

Thats my opinion

evry1cndie

#### evry1cndie

What kind of integral is defined on an open interval?

I have no idea, unless this was a rhetorical question.

This isn't true for the Riemann integral, since we can extend it to a function in [0,1] as 0 in the endpoints, and it would only be discontinous there.

this was my thought exactly, then if it can be extended to the closed interval it can be integrated and then integrable on all sub-intervals including the open one...

I dont think it is true either, if f is continuous and bounded over (0,1) then

$$\displaystyle \bigg|\int_0^1 f(x)dx\bigg|\leq\bigg|\int_{(0,1)} fd\lambda\bigg|\leq M \lambda((0,1))<\infty$$

so f would be integrable.

Thus it is important to remove some of the hypothesis:

continuity or boundedness or measure finite set

so we can have some functions satisfying the statement that is left after the removal.

Thats my opinion

I'm unfamiliar with this notation, what's lambda in this case? just a change of variable?

Last edited:

#### mabruka

$$\displaystyle \lambda$$ denotes the Lebesgue measure , nevermind then,

the property im using in terms of Riemann integrals :

if f is continuous and bounded then

$$\displaystyle \bigg |\int_a^b f(x)dx \bigg|\leq \int_a^b |f(x)|dx\leq \ (b-a)M$$

So if f is bounded then you can bound the integral by f's bound multiplying it by the lenght of the integrating interval.

This is a consequence of integral monotonicity, this says:

if $$\displaystyle g(x) \leq h(x)$$ for all x then

$$\displaystyle \int g(x) dx \leq \int h(x) dx$$

Using the above for $$\displaystyle |f(x)|\leq M$$ we get the first one.

evry1cndie

#### Drexel28

MHF Hall of Honor

#### TheEmptySet

MHF Hall of Honor
Question: Give an Example of a Function Which is Continuous and Bounded on the Open Interval (0,1) and prove it....

The proof i hopefully would be able to do. I have Absolutely no idea What kind of function would be non-integrable.

The fact that it's bounded and continuous almost seems to guarantee the functions integrability, the only thing i see destroying it is the open interval, however looking at it in the sense if Darboux Upper/Lower Sums, Sup{f(x)} and Inf{f(x)} need not belong to the interval, so even if the function achieves a max/min at the endpoints and not within the interval we can still integrate... maybe I'm thinking too hard.. i dunno. help!
The function

$$\displaystyle f(x)=\sin\left( \frac{1}{x}\right)$$
is continous on $$\displaystyle (0,1)$$ and bounded there but is not integrable and cannot be continously extented to 0.

#### Drexel28

MHF Hall of Honor
The function

$$\displaystyle f(x)=\sin\left( \frac{1}{x}\right)$$
is continous on $$\displaystyle (0,1)$$ and bounded there but is not integrable and cannot be continously extented to 0.
So? Define $$\displaystyle f(0)=0,f(1)\sin(1)$$. This is not continuous but surely integrable.

#### evry1cndie

.... ok, before everyone starts fighting like dogs, the question was improperly phrased. it should have been "prove/disprove."(Headbang) spoke to my professor today... Thanks all for you're help. I was thinking of the sin(1/x) but also found it was integrable, the only second option was a sawtooth function with base 1/n... but it faced the same fate as sin(1/x) it would surely be integrable... Thanks for you're help guys... sorry if this question made you doubt the integrability of continuous functions. heh.