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
InfU(p,f)=supL(p,f)Inf U (p, f) =sup L(p, f)InfU(p,f)=supL(p,f)
L(p,f)=∑(i=1)miΔxiL(p, f) =\sum_(i=1) mi\Delta xiL(p,f)=∑(i=1)miΔxi
L(p,f)=m1Δx1+m2Δx2+...+mrΔxr+...+mnΔxnL(p, f) =m1\Delta x1+m2\Delta x2 +... +mr\Delta xr +... +mn\Delta xnL(p,f)=m1Δx1+m2Δx2+...+mrΔxr+...+mnΔxn
mi=0∴L(p,f)=0mi=0 \therefore L(p, f) =0mi=0∴L(p,f)=0
SupL(p,f)=0Sup L (p, f) =0SupL(p,f)=0
U(p,f)=∑(i=1)MiΔxiU(p, f) = \sum _(i=1) Mi\Delta xiU(p,f)=∑(i=1)MiΔxi
U(p,f)=M1Δx1+M2Δx2+...+MrΔxr+...+MnΔxnU(p, f) =M1\Delta x1+M2\Delta x2+... +Mr\Delta xr+... +Mn\Delta xnU(p,f)=M1Δx1+M2Δx2+...+MrΔxr+...+MnΔxn
M1=0,Mr=1M1=0, Mr =1M1=0,Mr=1
U(p,f)=1ΔxrU(p, f) = 1\Delta xrU(p,f)=1Δxr
Δxr→0∴infU(p,f)=0\Delta xr\to 0 \therefore inf U(p, f) =0Δxr→0∴infU(p,f)=0
Hence integrable
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