Increasing sequence of step functions and diverging Lebesgue Integrals

Question

Let $\{s_n\}$ be an increasing sequence of step functions which converges pointwise on an interval $I$ to a limit function $f$. If $I$ is unbounded and if $f(x) \geq 1$ everywhere on $I$ except on a set of measure zero, prove that the sequence $\{\int_I s_n\}$ diverges. (Here $\int_I s_n$ refers to the Lebesgue Integral.)

Conceptually, this makes sense. Clearly, if $f(x) \geq 1$ almost everywhere on an unbounded interval, then the sequence of Lebesgue integrals should diverge. However, in proving this, I am running into issues with the convergence of $\{s_n\}$ being pointwise. If it were uniform, I could easily show this.

How do I show that (even for pointwise convergence) I can choose $n$ large enough to make $\int_I s_n$ as big as I want?

Answer 1

I take step functions mean at least that the image of $s_n$ is finite and $\{s_n \ne 0\}$ is finite measured. In particular, $s_n$ is integrable.

Assume $s_1^- := \max(-s_1,0)$ has positive integral. Otherwise we can apply monotone convergence theorem directly.

Notice that $\int_I s_1^-$ is finite, as $s_1$ is a step function and thus integrable. Then, $s_n + s_1^- \ge s_1 + s_1^- = s_1^+ = \max(s_1,0)$ is non-negative and increases pointwise to $f - s_1^-$. Thus, MCT applies and yields $$ \lim_{n\to\infty} \int_I s_n + s_1^- = \int_I f + s_1^- \ge \int_I f = \infty. $$ Thus, it follows $$ \int_I s_n = \int_I s_n + s_1^- - \underbrace{\int_I s_1^-}_{< \infty} \to \infty. $$

Aside: Notice that $s_n$ is a step function is not crucial. We can replace it with "$s_n$ is integrable".

Answer 2

Does this work?:

We have that $s_n$ converges from below to $f$ almost everywhere on I.

Assume $\lim_{n\rightarrow \infty} \int_I s_n$ is finite. Then $f$ is an upper function, denoted $f \in U(I)$.

Then we have that $\lim_{n\rightarrow \infty} \int_I s_n = \int_I f$.

However, $f \geq 1$ a.e. on $I$. So $\int_I f\geq \int_I 1 = 1 \cdot \mu(I) = \infty$.

Therefore $f \notin U(I)$ and $\int_I s_n$ diverges.

Answer 3

Let $f_n =s_n\chi_{s_n^{-1}((0,\infty)))}$. Then, $f_n\geq 0$ are measureable, and increase to $f$ a.e.:

Let $x\in I$. Since $s_n(x)\nearrow f(x)\geq 1$ we can find an $N\in \mathbb N$ s.t. $n>\mathbb N\Rightarrow s_n(x)>0$. But then, $s_n(x)=f_n(x)$.

Now use MCT.