This exercise is the other one that I have some trouble with.

Some usual definitions can be acquired here: Questions of an exercise in Lebesgue integral, (obviously, that was the first one which I got stuck by).

Another definition probably used, Dominated Convergence Theorem: Suppose $f_k \in L(E) (k=1,2,3,...)$ where $L(E)$ means a set of measurable functions defined on $E \subset \mathbb R^n$ while each function $f$ in it satisfies $||f||_{1} = \int_{E}|f(x)|dx < \infty$(Note: it is a Lebesgue integral). And $$\lim_{k \to \infty} f_k(x) = f(x), a.e. x \in E.$$ If $\exists F(x)$, a Lebesgue integrable function defined on $E \subset \mathbb R^n$ such that $$|f_k(x)| \le F(x), a.e. x \in E \ (k=1,2,3,...),$$ then $$\lim_{k \to \infty} \int_{E} f_k(x)dx = \int_{E}f(x)dx.$$ Ususally, we call $F(x)$ is a dominated function.

Exercise2: Suppose $f \in L(E)$ and $f(x) > 0( \forall x \in E)$ then show $$\lim_{k \to \infty} \int_{E}(f(x))^{\frac{1}{k}}dx = m(E).$$

My trial is:Let $A = \{x \in E: f(x) < 1 \}$ then $$\lim_{k \to \infty} \int_{E}f^{\frac{1}{k}}(x)dx = \lim_{k \to \infty} \int_{A}f^{\frac{1}{k}}(x)dx + \lim_{k \to \infty} \int_{E-A}f^{\frac{1}{k}}(x)dx.$$ Then I got stuck here coz I did't know how to deal with limit in front of an integral.

Solution: Go on doing what I have done above, that is $$\lim_{k \to \infty} \int_{A}f^{\frac{1}{k}}(x)dx + \lim_{k \to \infty} \int_{E-A}f^{\frac{1}{k}}(x)dx = \int_{A} \lim_{k \to \infty} f^{\frac{1}{k}}(x)dx + \int_{E-A} \lim_{k \to \infty} f^{\frac{1}{k}}(x)dx = \int_{A} 1 dx + \int_{E-A}1dx = m(E).$$ So I think probably it has taken dominated convergence theorem into effect. However, what are the two dominated functions for $f(x)$ on $A$ and $E-A$, respectively? Coz I think the exercise doesn't say $m(E) < +\infty.$

  • $\begingroup$ @Bungo: $1$ may not be Lebesgue integral on $A$ coz $m(A)$ may not be finite. Why $f_1$ on $E \setminus A$? $\endgroup$ Aug 7, 2015 at 6:09
  • $\begingroup$ $(f^{1/k})$ is dominated by the constant function $1$ on the set $A$, and by the function $f$ on the set $E \setminus A$, because if $f > 1$, the sequence $(f^{1/k})$ is decreasing. $\endgroup$
    – user169852
    Aug 7, 2015 at 6:09
  • $\begingroup$ Right, if $m(A)$ is infinite, then so is $m(E)$, so you need to show that $\lim_{k \to \infty}\int_E (f(x))^{1/k} dx = \infty$ in that case. $\endgroup$
    – user169852
    Aug 7, 2015 at 6:14
  • $\begingroup$ @Bungo: well, why "you need to show that $\lim_{k \to \infty}(f(x))^{1/k} dx = \infty$"? $\endgroup$ Aug 7, 2015 at 6:18
  • $\begingroup$ @Bungo: So $m(A)$ must be finite? $\endgroup$ Aug 7, 2015 at 6:19

1 Answer 1


I'll use your notation, $A = \{x \in E : f(x) < 1\}$. First, note that $(f^{1/k})$ is increasing on $A$ and decreasing on $E \setminus A$. Therefore, $(f^{1/k})$ is dominated by the constant function $1$ on $A$, and by the function $f^{1/1} = f$ on $E \setminus A$. Also, of course, $f^{1/k} \to 1$ on both $A$ and $E \setminus A$.

Now consider two cases.

Case 1: $m(A) < \infty$

In this case, the constant function $1$ is integrable on $A$, so by the dominated convergence theorem, $$\lim_{k \to \infty} \int_A f^{1/k} = \int_A \lim_{k \to \infty} f^{1/k} = \int_A 1 = m(A)$$ Since $f$ is integrable on $E$, it is also integrable on $E \setminus A$, so again by the DCT, the limit and integral can also be interchanged on $E \setminus A$, yielding the desired result.

Case 2: $m(A) = \infty$

In this case, $m(E)$ is also infinite since $A \subseteq E$. Therefore we need to show that $$\lim_{k \to \infty} \int_E f^{1/k} = \infty$$ and to do this, it suffices to show that $$\lim_{k \to \infty} \int_A f^{1/k} = \infty$$ since $f$ is nonnegative. To do this, we recall that $(f^{1/k})$ is increasing on $A$, so the monotone convergence theorem applies and we can interchange the limit and integral: $$\lim_{k \to \infty} \int_A f^{1/k} = \int_A \lim_{k \to \infty} f^{1/k} = \int_A 1 = m(A) = \infty$$

  • $\begingroup$ That's clear. Thanks Bungo^_^. I will up vote it. $\endgroup$ Aug 7, 2015 at 6:29
  • $\begingroup$ @Bearandbunny: Note that we can actually combine the two cases by always using the monotone convergence theorem on $A$, and the dominated convergence theorem on $E \setminus A$. $\endgroup$
    – user169852
    Aug 7, 2015 at 6:33

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