# Lebesgue Measure Space and a converging, increasing sequence. Show that the sequence in increasing.

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.

• Hint: use the monotone convergence theorem.
– Hugo
Nov 21, 2019 at 15:25
• 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. Nov 21, 2019 at 15:35
• You apply the monotone converge theorem to the functions $f_n = f \cdot \chi_{(\frac{1}{n}, 1]}$ (which are Lebesgue measurable).
– Hugo
Nov 21, 2019 at 15:39
• So this could be rewritten as $\int _{(0,1]} fd\lambda = \int_{(\frac1n,1]} \lim n \to ∞ f_n d\lambda$ Nov 21, 2019 at 15:56

• 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.
• 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? Nov 21, 2019 at 18:17
• @martinhynesone It is correct
– user515010
Nov 21, 2019 at 18:21