Question on proof of Riemann integrability of $(\alpha\cdot f)$ Let be $f$ a Riemann integrable function on $[a,b]$ and $\alpha<0$. Then, by definition of Riemann integrability
$$
\inf\{U(f,P)\mid P\text{ is partition of }[a,b]\}=\sup\{L(f,P)\mid P\text{ is partition of }[a,b]\},
$$
where $U(f,P)$ and $L(f,P)$ are upper and lower Darboux sums. It is pretty standard to show that $(\alpha\cdot f)$ is also Riemann-integrable with $\int\limits_a^b\alpha\cdot f(x)dx=\alpha\cdot \int\limits_a^bf(x)dx$.
However, I had a little discussion with my tutor on the following proof where she said that it can't be done this way and I should stick to the "official proof" from lecture. Oddly enough she was unable to explain to me where and why it breaks:

As $\alpha<0$, we conclude
\begin{align*}
&\inf\{U(f,P)\mid P\text{ is partition of }[a,b]\}=\sup\{L(f,P)\mid P\text{ is partition of }[a,b]\}\\
&\iff\\
&\alpha\inf\{U(f,P)\mid P\text{ is partition of }[a,b]\}=\alpha\sup\{L(f,P)\mid P\text{ is partition of }[a,b]\}\\
&\iff\\
&\sup\{\alpha U(f,P)\mid P\text{ is partition of }[a,b]\}=\inf\{\alpha L(f,P)\mid P\text{ is partition of }[a,b]\}\\
&\iff\\
&\sup\{ L(\alpha\cdot f,P)\mid P\text{ is partition of }[a,b]\}=\inf\{U(\alpha\cdot f,P)\mid P\text{ is partition of }[a,b]\}\\
&\\
&\implies (\alpha\cdot f)\text{ is Riemann-integrable}.
\end{align*}
I have used the known facts that that $\sup$ turns into $\inf$ (and vice versa) if the elemnts of the set are multiplied with a neagtive real and that each $U(f,P)$ and $L(f,P)$ is simply finite sum of $\sup$'s and $\inf$'s.

What is wrong with this approach?
 A: This is a valid approach based on the properties $$\sup\{ -a: a \in S\}= - \inf\{a: a \in S\},\quad \inf\{ -a: a \in S\}= - \sup\{a: a \in S\},$$
and, for $c > 0$,
$$\sup\{ ca: a \in S\}= c \sup\{a: a \in S\},\quad \inf\{ ca: a \in S\}= c \inf\{a: a \in S\}$$
To provide more justification, we have from the basic properties when $\alpha < 0$,
$$\alpha \inf_PU(f,P) =-|\alpha|\inf_P U(f,P) = |\alpha|\sup_P[-U(f,P)] = \sup_P[-|\alpha|U(f,P)]= \sup_P[\alpha U(f,P)],$$
and by a similar argument
$$\alpha \sup_PL(f,P)= \inf_P [\alpha L(f,P)]$$
This justifies your first two if-and-only-if statements.
To justify your third if-and-only-if statement, note that for any partition interval $I$, we have
$$\sup_{x \in I} [\alpha f(x)]= \sup_{x \in I} [-|\alpha| f(x)]= -\inf_{x \in I}[|\alpha|f(x)]= -|\alpha|\inf_{x \in I} f(x)= \alpha \inf_{x \in I} f(x),$$
which implies that $U(\alpha f,P) = \alpha L(f,P)$. By a similar argument we can show that $L(\alpha f,P) = \alpha U(f,P)$.
Thus,
$$\sup_P [\alpha U(f,P)] = \sup_PL(\alpha f,P), \quad \inf_P [\alpha L(f,P)] = \inf_P U(\alpha f,P)$$
