Proof that the Riemann-Integral satisfies $\int_A \lambda f = \lambda \int_A f$ Suppose $A\subset\mathbb{R}^n$ is a closed rectangle and $f:A\to \mathbb{R}$ is Riemann-Integrable on $A$. I want to show that $\lambda f$ is integrable and that
$$\int_A \lambda f =\lambda\int_Af $$
My approach was the following: suppose $\lambda \geq 0$, by integrability of $f$ we know that there is a partition $P$ of $A$ such that $U(f,P)-L(f,P)< \epsilon/\lambda$. Then, let $S$ be a subrectangle of this partition, we know that $m_S(f)\leq f(x)\leq M_S(f)$ for all $x\in S$. Since $\lambda \geq 0$ we have $\lambda m_S(f) \leq \lambda f(x)\leq M_S(f)$ for all $x \in S$, and hence, since $m_S(\lambda f)$ is the greatest lower bound and $M_S(\lambda f)$ is the least upper bound we have $m_S(\lambda f) \geq \lambda m_S(f)$ and $M_S(\lambda f)\leq \lambda M_S(f)$.
In this case, we have the relation for the upper and lower sums:
$$U(\lambda f,P) =\sum_{S\in P}M_S(\lambda f,P)v(S) \leq \lambda \sum_{S\in P}M_S(f)v(S) = \lambda U(f,P)$$
$$L(\lambda f,P)= \sum_{S\in P}m_S(\lambda f,P)v(S) \geq \lambda \sum{S\in P}m_S(f)v(S) = \lambda L(f,P)$$
and this gives us $U(\lambda f,P)-L(\lambda f,P) \leq \lambda(U(f,P)-L(f,P)) < \lambda \epsilon/\lambda = \epsilon$ which proves integrability.
In the case $\lambda < 0$, there is a partition $P$ of $A$ such that $U(f,P)-L(f,P)< -\epsilon/\lambda$. Now, if $S$ is a subrectangle of $P$, from $m_S(f)\leq f(x)\leq M_S(f)$ for all $x\in S$ we have $\lambda M_S(f) \leq \lambda f(x) \leq \lambda m_S(f)$ for all $x\in S$. Hence, by similar arguments as above, $m_S(\lambda f)\geq \lambda M_S(f)$ and also $M_S(\lambda f)\leq \lambda m_S(f)$. This gives us
$$U(\lambda f,P)=\sum_{S\in P}M_S(\lambda f,P)v(S) \leq \lambda \sum_{S\in P}m_S(f)v(S) = \lambda L(f,P)$$
$$L(\lambda f,P)= \sum_{S\in P}m_S(\lambda f,P)v(S) \geq \lambda \sum_{S\in P}M_S(f)v(S) = \lambda U(f,P)$$
and this gives $U(\lambda f,P)-L(\lambda f,P) \leq\lambda(L(f,P)-U(f,P)) < \lambda \epsilon/\lambda = \epsilon$ and therefore $\lambda f$ is integrable again.
Now, first of all: is this proof correct? I think it is correct, but I'm a little unsure on the case $\lambda < 0$. Also, I'm stuck now, I need a hint on how to show that
$$\int_A \lambda f = \lambda \int_A f.$$
Thanks very much in advance!
 A: Cleaned-up Version:
Let $\lambda \ne 0, \epsilon > 0$ be given and chose a partition $P$ of $A$ s.t.
$$U(f;P) - L(f;P) < \frac \epsilon {|\lambda|}$$
WLOG assume $\lambda > 0$ since $f$ is Riemann integrable on $A \Leftrightarrow -f$ is Riemann integrable on $A$ (shown?)
Then see that $\def\vol{\ {\rm vol}}$
$$U(\lambda f;P) = \sum_{P_i \in P} \max_{x\in P_i} \lambda f(x) \vol(P_i) = \sum_{P_i \in P} \lambda \max_{x\in P_i} f(x) \vol(P_i) = \lambda \sum_{P_i\in P} \max_{x\in P_i} f(x) \vol(P_i) = \lambda U(f; P)$$
$$L(\lambda f;P) = \sum_{P_i \in P} \min_{x\in P_i} \lambda f(x) \vol(P_i) = \sum_{P_i \in P} \lambda \min_{x\in P_i} f(x) \vol(P_i) = \lambda \sum_{P_i\in P} \min_{x\in P_i} f(x) \vol(P_i) = \lambda L(f; P)$$
So now
$$U(\lambda f; P) - L(\lambda f; P) = \lambda U(f;P) - \lambda L(f;P) = \lambda (U(f;P) - L(f;P)) < \lambda \frac\epsilon\lambda = \epsilon$$
There for $\lambda f$ is Riemann integrable and
$$\int_A \lambda f = \lim_{|P| \to\infty} U(\lambda f;P) = \lim_{|P|\to\infty} \lambda U(f;P) = \lambda \lim_{|P|\to\infty} U(f;P) = \lambda \int_A f$$
(the last part can also be formulated with $L(f;P)$ etc.)
