$f,g:P \rightarrow \mathbb{R}$ bounded on P are Riemann integrable. Then $f \cdot g$ is Riemann integrable.
I have above proof in my lecture : \begin{align} |(gf)(x)-(gf)(y)|&=|g(x)f(x)-g(y)f(y)|\\ &=|g(x)f(x)-g(x)f(y)+g(x)f(y)-g(y)f(y)|\\ &\le|g(x)||f(x)-f(y)|+|f(y)||g(x)-g(y)|\\ &\le A|f(x)-f(y)|+B|g(x)-g(y)| \end{align} where $A,B$ are constants that are bounding functions $f$ and $g$. And now I have written that $fg$ is integrable, but I don't see why. Can you please explain me why we have such a corollary ?
Let $\Delta x_k$ denote the length of the partition's $k$-th interval $\Delta_k$.
We know that $$f\text{ is Riemann-integrable}\iff\sum_k\sup_{x,y\in\Delta_k}(|f(x)-f(y)|)\cdot\Delta x_k\to0$$ as the mesh of the partition $\Delta$ goes to $0$.
So, in this case, we have $$|(fg)(x)-(fg)(y)|\leq A|f(x)-f(y)|+B|g(x)-g(y)|$$ $$\leq A\sup_{x,y\in\Delta_k}|f(x)-f(y)|+B\sup_{x,y\in\Delta_k}|g(x)-g(y)|$$
After building the sum over all intervals of the partition, this means that $$\sum_k\sup_{x,y\in\Delta_k}|(fg)(x)-(fg)(y)|\leq A\sum_k\sup_{x,y\in\Delta_k}|f(x)-f(y)|\Delta x_k+B\sum_k\sup_{x,y\in\Delta_k}|g(x)-g(y)|\Delta x_k.$$
Now, both $\sum_k\sup_{x,y\in\Delta_k}|f(x)-f(y)|\Delta x_k$ and $\sum_k\sup_{x,y\in\Delta_k}|g(x)-g(y)|\Delta x_k$ go to $0$ as the partition's mesh goes to $0$ because $f$ and $g$ are Riemann-integrable, and so does the whole term. Thus, $fg$ is Riemann-integrable.