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.

  • $\begingroup$ If we're talking about the Lebesgue integral, $\displaystyle\int_0^1 f$ exists for the first example since it constant almost everywhere. Have you studied the dominated convergence theorem? $\endgroup$
    – user17794
    Dec 18, 2013 at 6:53
  • $\begingroup$ I do not know dominated convergence theorem. $\endgroup$
    – Supriyo
    Dec 18, 2013 at 6:56
  • $\begingroup$ The theorem should be discussed in any text on measure theory. You'll probably see uniform boundedness replaced with pointwise boundedness by a non-negatative integrable function - I don't know of a cleaner proof for the special case in 4). $\endgroup$
    – user17794
    Dec 18, 2013 at 7:04
  • $\begingroup$ @TimDuff I do not know measure theory. Please just tell me which topics should I know to solve the problem completely? domineted convergence theorem, pointwies boundedness, uniform boundedness and then? $\endgroup$
    – Supriyo
    Dec 18, 2013 at 10:18

1 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.

  • $\begingroup$ Thank you Euler for your interest to my question. So where is wrong with my example/ $\endgroup$
    – Supriyo
    Dec 18, 2013 at 6:57
  • $\begingroup$ What is wrong with the example is that in fact the integral exists and is equal to $0$ $\endgroup$ Dec 18, 2013 at 6:58

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