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The question is as following, by using Lebesgue integral definition, show that $\int_{[0,1]}fdm=\frac{1}{2}$ where $f(x) = x$ on $[0,1]$. Here is my approach.

For any $\epsilon>0$ and $A=[0,1]$, choose $\frac{1}{2}<\delta<\frac{\epsilon}{m(A)}$ such that whenever $||P||<\delta$, we have

$$|L(f,P)-\frac{1}{2}|=|\sum_{i=1}^{n}y_i^*.m(\{x\in A|y_{i-1}\leq f(x)<y_i\})-\frac{1}{2}\sum_{i=1}^{n}m(\{x\in A|y_{i-1}\leq f(x)<y_i\})|=|\sum_{i=1}^{n}(y_i^*-\frac{1}{2}).m(\{x\in A|y_{i-1}\leq f(x)<y_i\})|=\sum_{i=1}^{n}|x_i^*-\frac{1}{2}|.m(\{x\in A|y_{i-1}\leq f(x)<y_i\})<\sum_{i=1}^{n}\frac{1}{2}.m(\{x\in A|y_{i-1}\leq f(x)<y_i\})<\delta.\sum_{i=1}^{n}.m(\{x\in A|y_{i-1}\leq f(x)<y_i\})=\delta.m(A)<\epsilon$$ as required.

Am I correct with this proof ?

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  • $\begingroup$ You have proved Riemann integrability. I haven't seen any body defining Lebesgue integral in this fashion. You have used the tags measure theory and lebegue-integral but have you studied Lebesgue intargation? $\endgroup$ – Kavi Rama Murthy Jun 25 at 5:53
  • $\begingroup$ Isn't this Lebesgue integration ?? I used measurable set to prove it . $\endgroup$ – Ling Min Hao Jun 25 at 12:48

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