A few questions regarding using Markov/Chebyshev and writing the Lebesgue measure in terms of the Lebesgue Integral

Question

I just asked a question here Prove that $m(\{x\in[0,1]:\lim \sup_{j\rightarrow\infty}f_j(x)\geq\frac{1}{2}\})\geq\frac{1}{2}$ under these conditions..., and I am having a hard time seeing the bounds on the integral. I know it is dealing with Markov/Chebyshev, but I am just not completely seeing it. So, I wanted to ask the following:

1. In the question above, how exactly are the bounds determined? In particular, how do the sets $A$ and $B$ determine the bounds of the integral by generating the factor in front of $m(A)$ and $m(B)$, respectively? I assume it is just Markov, but it appears as the bounds on $f_j(x)$ is also playing a role I am not clearly seeing.
2. Suppose, for instance, $0\leq f_j(x)\leq 1$ instead of the bounds in the original problem... how would the value (bounds in terms of the measure, I suppose I should say) of the integral change?
3. In the sets $A$ and $B$, what if $\lim\inf$ was used instead of $\lim\sup$? Would that matter, or would that only make using Fatou for $B$ a bit easier since we wouldn't have to come up with equivalent statements for $\lim\sup$.
4. In the problem, $x\in[0,1]$, but,$f:[0,1]\rightarrow\mathbb{R}$. If instead of $\mathbb{R}$ we had some other interval, say $[0,1]$ (that is, say $f:[0,1]\rightarrow[0,1]$), how would that have affected the integral bounds in terms of the measure?

I suppose with these questions I am just trying to get a better sense of what I'm looking for and playing with when it comes to dealing with measurability problems. I would really appreciate any insight you all have! Thank you so much!

Answer 1

Question 1. If $f \leq g$ on a set $A$, then $\int_A f \ dm \leq \int_A g \ dm$. Furthermore, if $g = c$, where $c$ is a constant, then $\int_A g = c \ m(A)$. Thus, if $f \leq c$ on $A$, with $c$ constant, then $\int_A f \ dm \leq c \ m(A)$. That's all there is to it.

In the question that you linked, $f_n \leq \tfrac 3 2 $ for all $n$ everywhere (and in particular in $A$), so $\int_A f_n \ dm \leq \tfrac 3 2 m(A)$. Furthermore, $\limsup_{n \to \infty} f_n \leq \tfrac 1 2 $ on $B$, since $B$ is by definition the set of points where $\lim_{n \to \infty} f_n \leq \tfrac 1 2$, so $\int_B \limsup_{n \to \infty}f_n \ dm \leq \tfrac 1 2 m(B)$.

We're not using Markov or Chebyshev. (That said, Markov and Chebyshev are proved using very similar ideas...)

Question 2. If $0 \leq f_n (x) \leq 1$, then we have $\int_A f_n \leq 1 . m(A)$, so $$ 1 = \limsup_{n \to \infty} \int_0^1 f_n \ dm \leq \limsup_{n \to \infty} \int_A f_n \ dm + \limsup_{n \to \infty} \int_B f_n \ dm \leq m(A) + \tfrac 1 2 m(B) = \tfrac 1 2 m(A) + \tfrac 1 2 $$ Then $m(A) \geq 1$. But $A \subset [0, 1]$. So $m(A) = 1$.

Question 3. If we change $\limsup$ to $\liminf$ in the question but change nothing else, then the proof wouldn't work. For if $B := \{ x \in [0, 1] : \liminf_{n \to \infty} f_n(x) \leq \tfrac 1 2 \}$, then we get $$ \liminf_{n \to \infty} \int_B f_n \ dm \geq \int_B \liminf_{n \to \infty} f_n \ dm \leq \tfrac 1 2 m(B).$$ The $\geq $ inequality messes up the proof.

Question 4. In some sense, the $f_n$'s in the original problem can already be thought of as functions with bounded images. Since $0 \leq f_n \leq \tfrac 3 2$ by assumption, you can think of the $f_n$'s as functions $[0, 1] \to [0, \tfrac 3 2]$. Making the $f_n$'s functions to $[0, 1]$ is like saying that $0 \leq f_n \leq 1$ for all $x$. We've dealt with this scenario in Question 2.