I am trying to solve the following problem:
Let $I$ be an interval and $f: I \to \mathbb{R}$ be increasing. Show that $f$ is measurable by first showing that, for each natural number $n$, the strictly increasing function $x \mapsto f(x) + \frac{x}{n}$ is measurable, and then taking pointwise limits.
But I do not understand the function $x \mapsto f(x) + \frac{x}{n}$ is strictly increasing even thought there no restriction on $x$. Sorry if my question is stupid. Could someone help me in understanding this please?
Also, I know that if we define $f_n(x) = f(x) + \frac{x}{n}$ for each natural number $n$ and we fix a real number $c$ and define $B = \{x \in I: f_n(x) > c \}$ then, to solve the problem, we should show that $B$ is an interval. I feel like the Intermediate Value Theorem would help here but how? or just the strictly increasing property of $f_n(x)$? Any explanation will be greatly appreciated!
We are told that $f$ is (presumably weakly) increasing i.e. if $x<y$ then $f(x)\le f(y)$. Thus, if $x<y$ then $f(x)+\frac{x}{n}\le f(y)+\frac{x}{n}<f(y)+\frac{y}{n}$. So $x \mapsto f(x) + \frac{x}{n}$ is strictly increasing. More generally, if you add a weakly increasing function to a strictly increasing function then you will get a strictly increasing function (by the exact same steps).
We can just use the strictly increasing property. How might you tell that $B$ is an interval? Well, one sufficient condition is that if you pick any two points in $B$, every value between them is included (e.g. if $1,2\in B$, then $[1,2]\subseteq B$). So, let's pick $x,y\in B$ with $x<y$. Then, pick a value between them: $z$ satisfying $x<z<y$. We know that $f(x)>c$ and $z>x$ and $f_n$ is strictly increasing, so $f(z)>c$, so $z\in B$ as required.