# Converse of Lebesgue Monotone Convergence Theorem

As SquirtleSquad points out in the answers below (before I obnoxiously modified my question because I was hoping someone would find another example), some topology books present a "marching intervals" sequence of functions (term coined by John Kelley):

$f_1 = \begin{cases} 1 & x \in [0, \frac{1}{2}] \\ 0 & \text{otherwise} \end{cases}$

$f_2 = \begin{cases} 1 & x \in [\frac{1}{2}, 1] \\ 0 & \text{otherwise} \end{cases}$

$f_3 = \begin{cases} 1 & x \in [0, \frac{1}{4}] \\ 0 & \text{otherwise} \end{cases}$

$f_4 = \begin{cases} 1 & x \in [\frac{1}{4}, \frac{1}{2}] \\ 0 & \text{otherwise} \end{cases}$

$f_5 = \begin{cases} 1 & x \in [\frac{1}{2}, \frac{3}{4}] \\ 0 & \text{otherwise} \end{cases}$

Essentially, the "marching intervals," which have finite measure, shrink in size as $n \rightarrow \infty$, so $\int f_n \rightarrow 0$. However, $f_n$ does not converge to $0$ a.e. since the interval is always jumping around. It's the classic example that the converse of the monotone convergence theorem does not hold.

So my question is, is there another counterexample to the converse, or of a sequence of (Lebesgue) integrable functions $f_k: [a,b] \rightarrow [0,1]$ where $\int_{a}^{b} f_k \rightarrow 0$ as $k \rightarrow \infty$, but $f_k(x)$ does not converge almost everywhere to $0$?

Some functions that I tried: various functions involving the Cantor Set (and Fat Cantor Set) and manipulations with the characteristic function. Any suggestions or functions with no explicit formulas are also welcome. Thanks in advance.

Yes. Take $[a,b]=[0,1]$ and a sequence of functions as follows: The first is $1$ on the interval $[0,.5]$, and $0$ elsewhere. The second is $1$ on the interval $[.5,1]$, and $0$ elsewhere. The third is 1 on the interval $[0,.25]$, and $0$ elsewhere. The fourth is 1 on the interval $[.25,.5]$, and $0$ elsewhere, and so on. Then these functions do not converge pointwise, but their integrals converge to $0$.