- 176

- 0

hello all

i have been trying to prove a property of integrable functions, i had a go at it dont know if it is correct, but im wondering if there could possibly be a shorter simpler way of proving it alright here we go

[tex]\int_{a}^{b} f(x)+g(x) dx= \int_{a}^{b} f(x)dx +\int_{a}^{b} g(x) dx[/tex]

My proof

for any partition P of [a,b]

[tex]U(f+g,P)=\sum_{i=1}^{n}M_i(f+g,P)(x_i-x_{i-1})[/tex]

[tex]\le \sum_{i=1}^{n}M_i(f,P)(x_i-x_{i-1}) +\sum_{i=1}^{n}M_i(f,P)(x_i-x_{i-1}) [/tex]

[tex]= U(f,P)+U(f,g) [/tex] similarly

[tex]L(f+g,P)\ge L(f,P)+L(f,g)[/tex]

there is also partitions [tex] P_{1} [/tex] & [tex] P_{2} [/tex] of [a,b] such that

[tex] U(f,P_{1}) <\int_{a}^{b}f+\frac{\epsilon}{2} [/tex]

[tex] U(g,P_{2}) <\int_{a}^{b}g+\frac{\epsilon}{2} [/tex]

we let [tex] P=P_{1}UP_{2}[/tex]

[tex]\int_{a}^{b^U}(f+g)\le U(f+g,P)\le U(f,P)+U(g,P) \le U(f,P_{1})+U(g,P_{2})[/tex]

[tex]< \int_{a}^{b}f+\frac{\epsilon}{2}+\int_{a}^{b}g+\frac{\epsilon}{2}[/tex]

[tex]= \int_{a}^{b}f+\int_{a}^{b}g+\epsilon [/tex]

similarly

[tex]\int_{a_{L}}^{b}(f+g) > \int_{a}^{b}f+\int_{a}^{b}g -\epsilon [/tex]

since all functions are riemann integrable then

[tex]\int_{a}^{b^U}(f+g) =\int_{a_{L}}^{b}(f+g)= \int_{a}^{b}(f+g) [/tex]

and so it follows that

[tex]\mid\int_{a}^{b}(f+g)-\int_{a}^{b}f-\int_{a}^{b}g\mid<\epsilon[/tex] [tex]\forall\epsilon>0[/tex]

i have been trying to prove a property of integrable functions, i had a go at it dont know if it is correct, but im wondering if there could possibly be a shorter simpler way of proving it alright here we go

[tex]\int_{a}^{b} f(x)+g(x) dx= \int_{a}^{b} f(x)dx +\int_{a}^{b} g(x) dx[/tex]

My proof

for any partition P of [a,b]

[tex]U(f+g,P)=\sum_{i=1}^{n}M_i(f+g,P)(x_i-x_{i-1})[/tex]

[tex]\le \sum_{i=1}^{n}M_i(f,P)(x_i-x_{i-1}) +\sum_{i=1}^{n}M_i(f,P)(x_i-x_{i-1}) [/tex]

[tex]= U(f,P)+U(f,g) [/tex] similarly

[tex]L(f+g,P)\ge L(f,P)+L(f,g)[/tex]

there is also partitions [tex] P_{1} [/tex] & [tex] P_{2} [/tex] of [a,b] such that

[tex] U(f,P_{1}) <\int_{a}^{b}f+\frac{\epsilon}{2} [/tex]

[tex] U(g,P_{2}) <\int_{a}^{b}g+\frac{\epsilon}{2} [/tex]

we let [tex] P=P_{1}UP_{2}[/tex]

[tex]\int_{a}^{b^U}(f+g)\le U(f+g,P)\le U(f,P)+U(g,P) \le U(f,P_{1})+U(g,P_{2})[/tex]

[tex]< \int_{a}^{b}f+\frac{\epsilon}{2}+\int_{a}^{b}g+\frac{\epsilon}{2}[/tex]

[tex]= \int_{a}^{b}f+\int_{a}^{b}g+\epsilon [/tex]

similarly

[tex]\int_{a_{L}}^{b}(f+g) > \int_{a}^{b}f+\int_{a}^{b}g -\epsilon [/tex]

since all functions are riemann integrable then

[tex]\int_{a}^{b^U}(f+g) =\int_{a_{L}}^{b}(f+g)= \int_{a}^{b}(f+g) [/tex]

and so it follows that

[tex]\mid\int_{a}^{b}(f+g)-\int_{a}^{b}f-\int_{a}^{b}g\mid<\epsilon[/tex] [tex]\forall\epsilon>0[/tex]

Last edited: