the author is Halsey Royden , the title is "real analysis" fourth edition @KevinSong
let $f$ be a function of bounded variation on a closed and bounded interval $[a,b].$ Prove or disprove : $f$ is measurable.
Here is a wrong solution:
The statement is correct and here is the proof:\
let $f$ be a function of bounded variation on a closed and bounded interval $[a,b],$ then by Jordan's theorem on pg.117, $f$ is the difference of 2 increasing functions on $[a,b].$ Call them $h$ and $g,$ i.e. $f = h-g.$ \
Now, by thm.1 on pg. 108, each of $h$ and $g$ is continuous except possibly at a countable number of points in $(a,b).$ i.e. continuous a.e. on $(a,b).$i.e. discontinuity is on a set of measure 0 and in some cases this set of measure $0$ could be the $\emptyset.$\
Now, by proposition 3 on pg.55 , $h$ and $g$ are measurable. Note that in our case the measurable set upon which $f$ is defined and also $h$ and $g,$ is the closed interval $[a,b]$ and it is measurable by Proposition 8 on pg.38.\
Now, since the linear combination of 2 measurable functions is measurable by thm. 6 on pg.56 then the statement is proved.
the solution turns out to be wrong, could anyone tell me the correct solution please?
Best Answer
Monotone functions are themselves measurable, as their pre-images of intervals are again intervals.
For increasing $f$ and consider any $f^{-1}(a,\infty)$ and $x,y\in f^{-1}(a,\infty)$, let's assume that $x\leq y$ and so $a<f(x)\leq f(y)$. For any $t\in[0,1]$, we have $x\leq tx+(1-t)y\leq y$ and hence $f(x)\leq f(tx+(1-t)y)\leq f(y)$, this shows that $f(tx+(1-t)y)>a$ and hence $tx+(1-t)y\in f^{-1}(a,\infty)$.
So $f^{-1}(a,\infty)$ is connected, we know that in $\mathbb{R}$ the connected sets are all intervals (including singletons of course).