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The full questions is this:

Let $f:[a,b]\longrightarrow \mathbb{R}$ be Riemann integrable. Let $g:[a,b]\longrightarrow \mathbb{R}$ be a function such that $f(x)=g(x)$ for all $x\in (a,b)$. Prove that $g$ is Riemann integrable and that: $$\int_a^b g(x)\mathrm dx=\int_a^bf(x)\mathrm dx$$

I know that first I must prove that $g$ is Riemann integrable and I know to do that I must show the upper integral equals the lower integral. I also know that:

the lower integral is equal to $\sup\{L(P,g): P$ for all partitions of $ [a,b]\}$

and the upper integral is equal to $\inf\{U(P,g): P$ for all partitions of $ [a,b]\}$

Also, $$L(P,f)=\sum_{i=0}^n m_i\Delta x_i$$ where $m_i= \min\{f(x):\ x_{i-1}\leq x\leq x_i\}$ and $\Delta x_i=x_i-x_{i-1}$.

And $$U(P,f)=\sum_{i=0}^n M_i\Delta x_i$$ where $M_i=\max\{f(x):\ x_{i-1}\leq x\leq x_i\}$ and $\Delta x_i=x_i-x_{i-1}$.

I'm not quite sure how to go about doing this problem. I think I'm supposed to use the fact that $f(x)=g(x)$ but I'm a little stuck. Any help?

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  • $\begingroup$ As a hint, think what actually makes $g$ different from $f$. $\endgroup$ – xyzzyz Nov 15 '16 at 20:59
  • $\begingroup$ With Riemann sums. $\endgroup$ – hamam_Abdallah Nov 15 '16 at 21:09
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Hint

Let $P_n=\{x_0,\cdots, x_n\}$ be any partition of $[a,b].$ Then:

\begin{align} L(P,f)-m_0(f)(x_1-x_0)-m_n(f)(x_n-x_0)&\\=&\sum_{i=1}^{n-1}m_i(f)(x_{i+1}-x_i)\\=&\sum_{i=1}^{n-2}m_i(g)(x_{i+1}-x_i)\\=&L(P,g)-m_0(g)(x_1-x_0)-m_n(g)(x_n-x_0).\end{align}

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