Property of integral and integrator. (NET June 2011)

Question

Consider the sequence of integrable functions $f_n(x)$ defined on the closed bounded interval $[a,b]$. Which of the following four options are correct?

1. If $f_n(x) \rightarrow 0$ almost everywhere then $\int_a^b f_n(x) \rightarrow 0$

2. If $\int_a^b f_n(x) \rightarrow 0$ then $f_n(x) \rightarrow 0$ almost everywhere.

3. If $f_n(x) \rightarrow 0$ almost everywhere and each $f_n(x)$ is a bounded function then $\int_a^b f_n(x) \rightarrow 0$ .

4. If $f_n(x) \rightarrow 0$ almost everywhere and each $f_n(x)$ is an uniformly bounded function then $\int_a^b f_n(x) \rightarrow 0$ .

Consider the function $f_n(x)$, on the interval $[0,1]$. Let $\{r_1, r_2, \dots\}$ is the enumerarion of rationals on $[0,1]$.

$$f_n(x) = \begin{cases} 1 & \text{x = $r_1, r_2, \dots r_n$} \\ 0 & \text{elsewhere} \end{cases}$$

Thus $f_n(x) \rightarrow f(x)$, which is $0$ almost everywhere

$$f(x) = \begin{cases}1 & x \in \mathbb{Q} \\0 & x \in [0,1] - \mathbb{Q}\cap [0,1]\end{cases}$$

But $\int_0^1 f(x)$ does not exists. Again eacu $f_n$ and $f$ are bouded and uniformly bounded. So 1,3,4, are false.

Consider another sequence of functions $g_n(x) = \sin(x), x \in [0,2\pi]$. $\int_0^{2\pi}g_n(x) dx = 0 $ $\forall n$. So $\int_0^{2\pi}g_n(x) dx \rightarrow 0$ almost everywhere but $g_n(x) \nrightarrow 0$. So 2 is false.

So all 4 are false. it is not so.

Where is the mistake in these examples? Which options will be true and why?

Thank you for your help. Second half of the question may be discussed. If it is , please give me the link.

Answer 1

1 is definitely false. Think of a sequence of rectangles with decreasing width and increasing height

2 is false. The "typewriter sequence" is the standard counterexample.

3 is false, by the same example as 1.

4 is true (assuming that when you say each function is uniformly bounded, that means that there is a uniform bound for all the functions). This is the bounded convergence theorem.