# Convergence in measure

1. Jan 4, 2012

### Fredrik

Staff Emeritus
Not sure where to post about measure theory. None of the forums seems quite right.

Suppose that $(X,\Sigma,\mu)$ is a measure space. A sequence $\langle f_n\rangle_{n=1}^\infty$ of almost everywhere real-valued measurable functions on X is said to converge in measure to a measurable function f, if for all ε>0, $\mu(\{x\in X:|f_n(x)-f(x)|\geq\varepsilon\})\to 0$.

The book says that it's easily seen that if $\langle f_n\rangle$ converges in measure to two measurable functions f and g, then f=g almost everywhere. I don't see it.

I understand that I need to prove that $\mu(\{x\in X: |f(x)-g(x)|>0\})=0$, but I don't even see how to begin.

2. Jan 4, 2012

### micromass

Note that

$$|f(x)-g(x)|\leq |f(x)-f_n(x)|+|f_n(x)-g(x)|$$

So if $|f(x)-g(x)|\geq \varepsilon$, then also

$$|f(x)-f_n(x)|+|f_n(x)-g(x)|\geq \varepsilon$$.

But this must mean that either $|f(x)-f_n(x)|\geq \varepsilon/2$ or $|f_n(x)-g(x)|\geq \varepsilon/2$ (proceed by contradiction: assume that neither holds). So

$$\{x\in X~\vert~|f(x)-g(x)|\geq \varepsilon\}\subseteq \{x\in X~\vert~|f(x)-f_n(x)|\geq \varepsilon/2\}\cup \{x\in X~\vert~|g(x)-f_n(x)|\geq \varepsilon/2\}$$

Taking $\mu$ of both sides yields

$$\mu\{x\in X~\vert~|f(x)-g(x)|\geq \varepsilon\}\leq \mu\{x\in X~\vert~|f(x)-f_n(x)|\geq \varepsilon/2\}+ \mu\{x\in X~\vert~|g(x)-f_n(x)|\geq \varepsilon/2\}$$

I think you can take it from here. It's just a limiting argument now.

Another nice argument (but a little more difficult to prove) is to show that the sequence has a subsequence which converges a.e. Now f=g a.e. by uniqueness of the limit a.e.

3. Jan 4, 2012

### I like Serena

From wikipedia:

"Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education."

and:

"Mathematical analysis, which mathematicians refer to simply as analysis, has its beginnings in the rigorous formulation of infinitesimal calculus. It is a branch of pure mathematics that includes the theories of differentiation, integration and measure, limits, infinite series,[1] and analytic functions."

Seems to me this fits Calculus & Analysis.

4. Jan 4, 2012

### micromass

I prefer this in the probability forum. Measure theory (and convergence in probability, like here) are really important probability concepts. So it might make sense that a probabilist knows more about them then an analyst (it certainly is my experience that this is the case).

5. Jan 4, 2012

### Fredrik

Staff Emeritus
Ah, I think I understand it now. Your argument tells us that for all $\varepsilon,\varepsilon'>0$, we have $\mu(\{x\in X:|f(x)-g(x)|\geq\varepsilon)<\varepsilon'$. This implies that for all $\varepsilon>0$, we have $\mu(\{x\in X:|f(x)-g(x)|\geq\varepsilon\})=0$. We still have to do something tricky for the final step, something like this:

$$\big\{x\in X:|f(x)-g(x)|>0\big\}=\bigcup_{n=1}^\infty\bigg\{x\in X:|f(x)-g(x)|\geq\frac{1}{n}\bigg\}$$
$$\mu\big(\big\{x\in X:|f(x)-g(x)|>0\big\}\big) \leq\sum_{n=1}^\infty\mu\bigg( \bigg\{x\in X:|f(x)-g(x)|\geq\frac{1}{n}\bigg\} \bigg)=0$$ Thank you.