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- Apr 13, 2013

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Let $f,g: [a,b] \to \mathbb{R}$ integrable functions.Show that: $\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g$

We suppose the subdivision $P=\{a=t_0<t_1<.....<t_n=b\}$ of $[a,b]$.

Let $t \in [t_k,t_{k+1}]$.

$$f(t) \leq sup f([t_k,t_{k+1}])$$

$$f(t) \geq inf f([t_k,t_{k+1}])$$

$$g(t) \leq sup g([t_k,t_{k+1}])$$

$$g(t) \geq inf g([t_k,t_{k+1}])$$

From these relations we get: $$u(f+g,P) \leq u(f,P)+u(g,P)$$

$$L(f+g,P) \geq L(f,P)+L(g,P)$$

where $L$ the lower sum and $U$ the upper sum.

As $f$ is integrable, $\forall \epsilon'>0 \exists P_1$ of $[a,b]$ such that $u(f,P_1)-L(f,P_1)<\epsilon'$.

As $g$ is integrable, $\forall \epsilon'>0 \exists P_2$ of $[a,b]$ such that $u(g,P_2)-L(g,P_2)<\epsilon'$.

We pick $P=P_1 \cup P_2$ and we get: $\int_{a}^{b}f+\int_{a}^{b}g< \epsilon + \underline{\int_{a}^{b}}(f+g)$

But why from this relation do we get: $\int_{a}^{b}f+\int_{a}^{b}g \leq \underline{\int_{a}^{b}}(f+g)$ ??

We also get: $\int_{a}^{b}f+\int_{a}^{b}g \geq \underline{\int_{a}^{b}}(f+g)- \epsilon$.And from this realtion,how do we get: $\int_{a}^{b}f+\int_{a}^{b}g \geq \underline{\int_{a}^{b}}(f+g)$ ?