Let $(R,L,\lambda)$ be the Lebesgue measure space and let $f: R → R$ be a non-negative, measurable function. Define a sequence $f_n =f.\chi _{(\frac1n,1]} : R→R, n \in N$

(a) Show that the sequence is increasing and converges pointwise to a function $f\cdot\chi_{(0,1]\$

(b) Show that $$\int_{(0,1]}f d\lambda = \lim_{n \to \infty} \int_{(\frac1n,1]}fd\lambda.$$

For (a), I just showed that at $n=1$, the region is $(1,1]$, at $n=2$, it is $(\frac12,1)$, and continues to infinity where the region is $(0,1)$, showing that the region is increasing, and converging. However, I am not very sure of how to use this to solve part b. Any help would be appreciated.

  • $\begingroup$ Hint: use the monotone convergence theorem. $\endgroup$
    – Hugo
    Nov 21, 2019 at 15:25
  • $\begingroup$ Can the monotone convergence theorem still be applied when the limit in this question is being applied to the boundary of integration? No examples I have been shown previously dealt with this sort of scenario. $\endgroup$ Nov 21, 2019 at 15:35
  • $\begingroup$ You apply the monotone converge theorem to the functions $f_n = f \cdot \chi_{(\frac{1}{n}, 1]}$ (which are Lebesgue measurable). $\endgroup$
    – Hugo
    Nov 21, 2019 at 15:39
  • $\begingroup$ So this could be rewritten as $\int _{(0,1]} fd\lambda = \int_{(\frac1n,1]} \lim n \to ∞ f_n d\lambda$ $\endgroup$ Nov 21, 2019 at 15:56

1 Answer 1

  • For (a) your approach works just fine. Alternatively, you could note that for $n>m$, $f_n-f_m=f\cdot\chi_{(\frac1n,\frac1m]}\ge 0$ and thus the sequence it's increasing. The pointwise convergence is then proven by $f-f_n=f\cdot \chi_{(0,\frac1n]}\to 0$.

  • For (b), after having written the problem as

$$\int_{(0,1]}f(x)\text{d}\mu(x)=\lim_{n\to \infty}\int f_n(x)\text{d}\mu(x)$$

you can either

  1. Use (a) and the monotone convergence theorem
  2. Use the pointwise convergence and the fact that $f_n\le f$ to apply the dominated convergence theorem.
  • $\begingroup$ Just to ensure my steps are right, after writing the problem as above, I left the left side of the equation untouched, and did the following steps to the r.h.s: $\int \lim_{n \to ∞} f_n(x)dμ(x)$, then $\int f_∞(x)dμ(x)$, and finally had it equal the l.h.s of the equation. Is this correct, or am I missing a step, or doing a step incorrectly? $\endgroup$ Nov 21, 2019 at 18:17
  • 1
    $\begingroup$ @martinhynesone It is correct $\endgroup$
    – user515010
    Nov 21, 2019 at 18:21

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