Confusion on Riemann integrable functions and further confusion about properties.

Question

Let $[a,b]$ an interval. Let $\Sigma$ the set of the subdivision of $[a,b]$. If $$\sigma :a=x_0<x_1<...<x_n=b$$ is a subdivision of $[a,b]$, we denote $$S_\sigma =\sum_{i=1}^n m_i(x_{i}-x_{i-1})\quad \text{and}\quad S^{\sigma }=\sum_{i=1}^n M_i(x_{i}-x_{i-1}),$$ where $$m_i=\min_{[x_{i-1},x_i]}f\quad \text{and}\quad M_i=\max_{[x_{i-1},x_i]}f.$$ I denote $$\overline{S}=\inf_{\sigma \in \Sigma} S^\sigma \quad \text{and}\quad \underline{S}=\sup_{\sigma \in \Sigma}S_\sigma .$$

Question 1 : Fo example, by definition of $\overline{S}$, can I say that there is a sequence $(\sigma _n)_{n\in\mathbb N}$ of $\Sigma$ s.t $$\overline{S}=\lim_{n\to \infty }S^{\sigma _n}\ \ ?$$

Question 2 : I have truble with the sequence $(\sigma _n)_{n\in\mathbb N}$. Can I say that such a sequence is decreasing in the sense that $\sigma _{n+1}\subset \sigma _n$ ? Or can I says that $(\sigma _n)_{n\in\mathbb N}$ is s.t. $h^{\sigma _{n+1}}\leq h^{\sigma _n}$ ? I recall that if $$\sigma _n : a=x_0<x_1<...< x_{n_1}=b$$ and $$\sigma _{n+1} : a=y_0<y_1<...<y_{n_2}=b,$$ then $\sigma _{n+1}\subset \sigma _n$ mean that for all $i=0,...,n_2-1$, there is $j\in \{0,...,n_1-1\}$ s.t. $[y_i,y_{i+1}]\subset [x_j,x_{j+1}]$ . And I denote $h^{\sigma _n}=\max\{|x_{i+1}-x_i|\ :\ i=0,...,n_{1}-1\}$ and same for $h^{\sigma _{n+1}}$.

Question 3 : With the definition inclusion for subdivision, can I say that $(\sigma \cup\tau)\subset \sigma $ and $(\sigma \cup\tau)\subset \tau$ ? Normally, for set it's the converse, but with subdivision, $\sigma \cup \tau$ looks to be the biggest subdivision $\kappa$ s.t. $\kappa \subset \sigma $ and $\kappa\subset \tau$. By the way, what would be $\sigma \cap \tau$ if it exist ?

Question 4 : Can I say that $f$ is Riemann integrable if and only if it's there is a sequence of step function $(f_n)$ s.t. $f_n(x)\to f(x)$ pointwise ? From my definition, the fact that Riemann integrable function are pointwise limit of step function looks straitforward. Is the converse true ?

Question 5: If Q4) is true, does it mean that $$\mathcal R[a,b]=Closure\{step\ function\}\ \ ?$$ where $\mathcal R[a,b]$ is the set of Riemann integrable function over $[a,b]$.

I'm sure that my questions are obvious for all of you, but I always have confusion with all theses things, so I would like to clarify all that.

Answer 1

Q1) Yes of course.

Q2) Yes you can suppose that $h^{\sigma _n}\to 0$.

Q4) Notice that you can define $$\underline{S}=\sup\left\{\int_a^b \varphi\mid \varphi\leq f,\ \varphi\ step\ function\right\},$$ $$\overline{S}=\inf\left\{\int_a^b \varphi\mid f\leq \varphi,\ \varphi\ step\ function\right\},$$ but if $f\in R[a,b]$ you won't have necessarily a sequence of step function $(f_n)$ s.t. $f_n(x)\to f(x)$ for all $x$ (it will be only almost everywhere).

Q5) Closure in which sense ? in $L^1$ ? Unfortunately, you will only have $$R[a,b]\subset Closure\{step\ function\}$$ since there are sequence of step function that converge to function that are not Riemann integrable (for example, there is a sequence of step function that will converge to $\boldsymbol 1_{\mathbb Q\cap [a,b]}$ that is not Riemann integrable.)