What Is an Example of a Nonintegrable Bounded Function?

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A nonintegrable bounded function example is the Dirichlet function, defined as f(x)=0 for irrational x and f(x)=1 for rational x in the interval (0,1). This function is bounded but not Riemann integrable due to its discontinuity at every point in the interval. While continuity is a sufficient condition for integrability, it is not necessary; a function can be integrable even if it is discontinuous on a set of measure zero. The discussion also touches on the distinction between Riemann and Lebesgue integrability, with examples illustrating each type. Understanding these concepts is crucial for calculus and analysis.
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could someone give me an example of a function that is bounded but is nonintegrable?


i need to know what a nonintegrable function bounded on [a,b] is as said in my preperation file for a test? urgent help needed
 
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How about
f:(0,1)\rightarrow R,

f(x)=0,x\in ((0,1) \cap R-Q)

f(x)=1,x\in ((0,1) \cap Q)

Daniel.
 
yup thnx, i forgot we did an example of dirichilet function in class, this function is an example of a lot of things used in calculus, lol,
 
Always remember that continuity is a necessary condition for integrability...

Daniel.
 
Continuity is a sufficient condition for integrability, not a necessary one. A function that is discontinuous on a set of points of measure 0 is integrable, and vice versa (i.e. this gives a necessary and sufficient condition). Clearly, a continuous function is discontinuous on an empty set which of course has measure 0, so it is integrable. The example you gave is discontinuous on (0, 1), a set that doesn't have measure 0, which is why f is not integrable. Of course, this also depends on how you define integration and integrability.
 
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Are u talking about Lebesgue,or Riemann integrability...?

Daniel.
 
dextercioby: The function f(x)= 0 if x< 0; 1 if 0< x< 1; 0 if x> 1 is (Riemann) integrable over any interval but is not continuous at 0 and 1.

The function: f(x)= 0 if x is rational; 1 if x is irrational is (Lebesque) integrable over any interval but is not continuous anywhere.
 
AKG's right, I'm pretty sure. IIRC, A bounded function is Riemann integrable over a compact set iff it's discontinuous on a set of measure zero.
 
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