@vv 

Please can you give me the two subsequences, 

@vv 

thank you

@vv 

The main goal: is if the group is nonabelian, if we compute the product ( using all possible order ), we can not get the always the identity.  
But, If the group is abelian, the product of all elements of G, is always equal to the identity. 

@sand15 

I agree with you, But how can I compute the sigma algebra generated by X 

@vv 

In our book, this theorem is named monotone class, but  the theorem must be stated as Synkin system ( which is the right definition) Thank you for your remark. 
How can we determine the element in this case under the definition of  Dynkin ystem generated by C

@vv 

Thank you.

@Kitonum 

Thank you for your lines code.

You code give a zero as result, but by hand we obtain 17 ( which is the right result)

@vv 

By hand we can compute the  Riemann Stieltjes integral  we obtain 17

@Zeineb 

Dear all

I tried to reproduce the same matlab working code, here, I get in this version 

Error, illegal use of an object as a name
 

The same steps are reproduced, but whta's wrong!!!

test3.mw

Thanks for your help 

@acer 

Thank you for your remarks, and sorry to post the same question twice. 
I hope someone look to my previous code and give me, what's wrong. Note that the Matlab code is a working code, and give me the correct values 

thanks

@Zeineb 

I hope receive any idea that solve the problem

@dharr 

thanks

@Zeineb 

http://www.ru.ac.bd/stat/wp-content/uploads/sites/25/2019/03/Burden_-Numerical-analysis-Thomson-Brooks_Cole-2005.pdf

@Zeineb 

The new code  work well ( no error )  but the code is also inplemented in MATLAB and gives two different results,  despite the same code lines used.
 

code_t.mw

thank you for any remarks that can solve this problem... two different results with same code 

here, is the MATLAB code 
 

f=@(t,y) y-t.^2+1;
TOL=10^(-9);
a=0;
b=2;
alpha=0.5
hmax=0.2;
hmin=0.01;
NK=[2 4 6 8 12 16 24 32];
TO=a;
WO=alpha;
h=hmax;
FLAG=1; % FLAG is used to exit from loop

for i=1:7
    for j=1:i
        Q(i,j)=(NK(i+1)/NK(j)).^2;
    end
end
fprintf('%6s %12s %18s %20s\n','to','wo','h','k');

while(FLAG==1)
    k=1;
    NFLAG=0;%Desired accuracy is achieved ,NFLAG is set to 1
    while(k<=8 && NFLAG==0)
        HK=h/NK(k);
        T=TO;
        W2=WO;
        W3=W2+HK*f(T,W2); % Eluer's First Step
        T=TO+HK;
     for j=1:NK(k)-1   
        W1=W2;
        W2=W3;
        W3=W1+2*HK*f(T,W2); %Mid point Method
        T=TO+(j+1)*HK;
      y(k)=(W3+W2+HK*f(T,W3))/2.0;   % end point correction to compute y(k,1)
    end
 
  if k>=2
      j=k;
      v=y(1);
      while(j>=2)
          y(j-1)=y(j)+( (y(j)-y(j-1)) /(Q(k-1,j-1)-1) );
          j=j-1;    
      end
    if(abs(y(1)-v)<=TOL)
        NFLAG=1;
    end
  end   
   k=k+1;
   end
k=k-1;
    if(NFLAG==0)
    h=h/2; %new value for w is rejected , decrease h
    end 
    if(h < hmin)
        FLAG=0;
    end
 WO=y(1);
TO=TO+h;
e=[TO WO h k];
fprintf('%6.2f        %8.14f       %1.2f                %1.2f\n',e);
if(TO>=b)
    FLAG=0;
elseif(TO+h > b)
        h=b-TO;
elseif(k<=3 && h< 0.5*(hmax))
    h=2*h;
end
hold on
plot(TO,WO,'p') %give us the plot of the numerical solution 
end 

@Carl Love 

Thank you for your remarks, using your remarks, the problem is solved. 

I appreciate your remarks. One again thanks