Prove that f = 0 if [tex]\int |f|= 0[\tex]

[mooberrymarz]

Guyz please help me !

hey! Could any of u please help me with this question. Its for my general integration theory assignment.

'Use an argumetn by contradiction to establish the following claim:
If f is continuous on R and \int |f|= 0,then f = 0.[\tex]<br /> <br /> thanx

 

[matt grime]

f continuous, |f| continuous, if f is not zero at some point, what can you say?

the *ml ends with /tex in the braces. avoid writing words in a latex environment: spaces do not get displayed as you type them and all letters are treated as variables and spaced accordingly.

 

[mooberrymarz]

ya

ok matt. wait I think i kinda get it. Would it be rite if i said that absolut evalue of f would not be equal to zero if f was not equal to zero. ( i think i might have to prove that somehow).Therefore the integral of the abs value would also not be equal to zero?

 

[matt grime]

that is what you must prove. what can you say about a continuous function in some neighbourhood of a point if it is not zero at that point?

sketch some positive function, not identically zero, on a bit of paper, pick some point where it's not zero, what can you see there? remember the integral is the area under the curve.

here's a nonexample:

if f is zero everywhere but x=1 where it is 1 then the integral over R is zero. that function is not continuous obviously, so where's the difference? look at the graphs

 

[Ebolamonk3y]

continous on its domain. not continuous everywhere... careful with limits of integration... Like 1/x for example... You find that's a funny function, one that converges as x approaches infinity but does NOT converge upon integration...

Integration is the difference in areas under or above the curves..

 

[matt grime]

Ebola, that doesn't address any points in the question.

 

[mooberrymarz]

got it. Thanx .

 

[matt grime]

would you mind running through your proof (if your teacher is pernickity it might be an idea to let someone check it).

 

[mooberrymarz]

haha. yes he is finicky. I used a graph to explain why integral of absolute value function would not be zero if f was not zero at any points. It kinda long and I don't want to type it all out. When i drew the graph it was obvious that the area under the graph of a positive function would be zero only if the funtcion was zero over a chosen interval.

I just am trying to put in proper mathematical english stuff. Umm, have i got the right idea? Am going to buy a crunchie bar.

 

[matt grime]

you have the idea, now you need to make it rigorous, when you've done that using epsilons and deltas and everything then you ought to post that proof (the proof is about 3 lines).

 

[mooberrymarz]

Epsilons and deltas?? Why? AHHHHH! Cant i just draw a graph and write stuff in plain english?

 

[matt grime]

because that is not a proof. if you're expected to learn lebesgue integration then this result is trivial, hence my puzzlement at your syllabus. I mean i can draw you a graph of a function that looks as though it is positive at lots of points and yet has integral zero. it's not continuous so you need to USE the continuity at some point.

 

[mooberrymarz]

really. interesting. Could u draw it for me while i sit and write out a epsilon delta proof for that question.

 

[matt grime]

this time let f be defined on the unit interval as 0 at irrational and 1 at rational. the lebesgue integral is zero, the function 'looks' like two parallel lines on its graph.

if you weren't planning an epsilon delta proof then i understand why your teacher knocks off marks, and it isn't being finicky.

 

[mooberrymarz]

Ok. Dont get angry... but why is the integral of that zero.?? Wouldnt u get many little dots, and summing their mini little areas under them would that not give some small number thingy??

 

[matt grime]

but it is non-zero on a set of measure zero only (the rationals). so the integral is zero. the integral cannot be done using riemann integration.

 

[mooberrymarz]

Would not riemann integrable imply not lebesgues integrable?

 

[matt grime]

it doesn't as that is equivalent ot lesbegue implies riemann, which is not true. that function is lebesgue integrable, as i said earlier.

 

[mooberrymarz]

thanx for your help!gotta go home and sleep so have a good day,k ;) cya