Question #189386

The function,f: [-1,1]×[2,1] → R, defined by

f(x,y)= { x ; y is rational


{ 0 ; y is irrational


is integrable or not.


True or false with full explanation


1
Expert's answer
2021-05-11T14:06:10-0400

InfU(p,f)=supL(p,f)Inf U (p, f) =sup L(p, f)

L(p,f)=(i=1)miΔxiL(p, f) =\sum_(i=1) mi\Delta xi

L(p,f)=m1Δx1+m2Δx2+...+mrΔxr+...+mnΔxnL(p, f) =m1\Delta x1+m2\Delta x2 +... +mr\Delta xr +... +mn\Delta xn

mi=0L(p,f)=0mi=0 \therefore L(p, f) =0

SupL(p,f)=0Sup L (p, f) =0


U(p,f)=(i=1)MiΔxiU(p, f) = \sum _(i=1) Mi\Delta xi

U(p,f)=M1Δx1+M2Δx2+...+MrΔxr+...+MnΔxnU(p, f) =M1\Delta x1+M2\Delta x2+... +Mr\Delta xr+... +Mn\Delta xn

M1=0,Mr=1M1=0, Mr =1

U(p,f)=1ΔxrU(p, f) = 1\Delta xr

Δxr0infU(p,f)=0\Delta xr\to 0 \therefore inf U(p, f) =0

Hence integrable


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