# Questions about Lebesgue integral of increasing measurable function converges in measure

I'm doing an exercise in Lebesgue integral.

Definitions probably used:

Some of them can be acquired here: Questions of an exercise in Lebesgue integral.

Definition of Convergence in measure: Suppose $f(x), f_1(x), f_2(x),..., f_k(x), ...$ are Lebesgue measurable functions that are finite almost everywhere ($m(\{x \in E: |f_k(x)| = +\infty\}) = 0$). $\{f_k(x)\}$ converge in measure to $f(x)$ if for $\forall \epsilon > 0,$ $\lim_{k \to \infty} m(\{ x\in E: |f_k(x) - f(x)| \ge \epsilon\}) = 0$.

Exercise3:

Plese note it is based on Lebesgue integral.

Suppose $0 \le f_1(x) \le f_2(x) \le ... \le f_k(x) \le ...(x \in E)$ and $f_k(x)$ converges in measurable to $f(x)$, then show $$\lim_{k \to \infty} \int_{E} f_k(x) dx = \int_{E} f(x) dx.$$

My trial and trouble:

There is no solution for this exercise. I tried to use Riesz's theorem of convergence in measure that is,

suppose $\{f_k(x)\}$ converge in measure to $f(x)$, then there exist a subsequence of $\{f_k(x)\}$, named $\{f_{k_i}(x)\}$ such that $\lim f_{k_i}(x) = f(x),$ almost everywhere, $x \in E.$

I can prove $\lim_{i \to \infty} \int_{E} f_{k_i}(x) dx = \int_{E} f(x) dx.$ However, I don't know how to show $\lim_{k \to \infty} \int_{E} f_k(x) dx = \lim_{i \to \infty} \int_{E} f_{k_i}(x) dx.$ The trouble is I can take $\lim$ into $\int_{E} f_{k_i}(x)$ by monotone convergence theorem that is $$\lim_{i \to \infty} \int_{E} f_{k_i}(x) dx = \int_{E} \lim_{i \to \infty} f_{k_i}(x) dx.$$ But I can't do that for $\lim_{k \to \infty} \int_{E} f_k(x) dx$ coz $\lim_{k \to \infty} f_k(x)$ may not exist.

• a monotone sequence of reals numbers has a accumulation point at most – user251257 Aug 8 '15 at 15:51
• For almost every $x\in E$, you have $f_{k_{i}}(x)\rightarrow f(x)$. $(f_{k}(x))$ is a monotone increasing sequence of real numbers, and $(f_{k_{i}}(x))$ converges to $f(x)$, and thus is bounded above by $f(x)$. It follows that $f(x)$ is also an upper bound for $(f_{k}(x))$, and so $(f_{k}(x))\rightarrow f(x)$. – catfish Aug 8 '15 at 16:06
• Yup, since $(f_{k_{i}}(x))\rightarrow f(x)$, $f(x)$ is a least upper bound for $(f_{{k}_{i}}(x))$, so it will have to be a least upper bound for $(f_{k}(x))$. – catfish Aug 8 '15 at 16:14
• In general, any convergent sequence in a metric space is bounded, so if $(f_{k_{i}}(x))$ consists of real numbers and converges, it is bounded in $\mathbb{R}$. It might be a little trickier here if we're working with the extended real line, $[-\infty,\infty]$ which is not a metric space on its own (it is homeomorphic to one however), but we're given that each $f_{k}$ is finite almost everywhere, and countable unions of measure 0 sets still have measure 0, so I think we should still be able to conclude via the monotone convergence theorem. Have you been able to solve it yet? – catfish Aug 8 '15 at 18:27
• I mean as topological spaces; both $[-\infty,\infty]$ and $[0,1]$ are topological spaces under the order topology, and $[-\infty,\infty]$ is homeomorphic to $[0,1]$. $[-\infty,\infty]$ is not a metric space but $[0,1]$ is (and the standard metric on $[0,1]$ induces the same topology as the order on $[0,1]$ does). This would allow us to consider the image of a sequence in $[-\infty,\infty]$ by the homemorphism which would then be a sequence in $[0,1]$ with the same topological properties. – catfish Aug 8 '15 at 19:26

## 1 Answer

For each $x\in E$ we have $\lim_{i\to\infty } f_{k_i} (x) =\lim_{k\to\infty} f_k (x)$ since $(f_n)$ is monotone. Analogously for the monotone sequence $a_k =\int_E f_k (t) dt$ we get $\lim_{i\to\infty} a_{k_i} =\lim_{k\to \infty } a_k .$