The Unapologetic Mathematician

Mathematics for the interested outsider

Convergence in Measure II

Sorry, I forgot to post this before I left this morning.

The proposition we started with yesterday shows us that on a set E of finite measure, a.e. convergence is equivalent to convergence in measure, and a sequence is Cauchy a.e. if and only if it’s Cauchy in measure. We can strengthen it slightly by removing the finiteness assumption, but changing from a.e. convergence to almost uniform convergence: almost uniform convergence implies convergence in measure. Indeed, if \{f_n\} converges to f almost uniformly then for any two positive real numbers \epsilon and \delta there is a measurable set F with \mu(F)<\delta and \lvert f_n(x)-f(x)\rvert<\epsilon for all x\in E\setminus F and sufficiently large n. Thus we can make the set F where f_n and f are separated as small as we like, as required by convergence in measure.

We also can show some common-sense facts about sequences converging and Cauchy in measure. First, if \{f_n\} converges in measure to f, then \{f_n\} is Cauchy in measure. We find that

\displaystyle\left\{x\in X\big\vert\lvert f_n(x)-f_m(x)\rvert\geq\epsilon\right\}\subseteq\left\{x\in X\bigg\vert\lvert f_n(x)-f(x)\rvert\geq\frac{\epsilon}{2}\right\}\cup\left\{x\in X\bigg\vert\lvert f_m(x)-f_m(x)\rvert\geq\frac{\epsilon}{2}\right\}

because if both f_n(x) and f_m(x) are within \frac{\epsilon}{2} of the same number f(x), then they’re surely within \epsilon of each other. And so if we have control on the measures of the sets on the right, we have control of the measure of the set on the left.

Secondly, if \{f_n\} also converges in measure to g, then it only makes sense that f and g should be “the same”. It wouldn’t do for a convergence method to have many limits for a convergent sequence. Of course, this being measure theory, “the same” means a.e. we have f(x)=g(x). But this uses almost the same relation:

\displaystyle\left\{x\in X\big\vert\lvert f(x)-g(x)\rvert\geq\epsilon\right\}\subseteq\left\{x\in X\bigg\vert\lvert f_n(x)-f(x)\rvert\geq\frac{\epsilon}{2}\right\}\cup\left\{x\in X\bigg\vert\lvert f_n(x)-g(x)\rvert\geq\frac{\epsilon}{2}\right\}

Since we can make each of the sets on the right arbitrarily small by choosing a large enough n, we must have \left\{x\in X\big\vert\lvert f(x)-g(x)\rvert\geq\epsilon\right\}=0 for every \epsilon>0; this implies that f=g almost everywhere.

Slightly deeper, if \{f_n\} is a sequence of measurable functions that is Cauchy in measure, then there is some subsequence \{f_{n_k}\} which is almost uniformly Cauchy. For every positive integer k we find some integer \bar{n}(k) so that if n,m\geq\bar{n}(k)

\displaystyle\mu\left(\left\{x\in X\bigg\vert\lvert f_n(x)-f_m(x)\rvert\geq\frac{1}{2^k}\right\}\right)<\frac{1}{2^k}

We define n_1=\bar{n}(1), and n_k to be the larger of n_{k-1}+1 or \bar{n}(k), to make sure that n_k is a strictly increasing sequence of natural numbers. We also define

\displaystyle E_k=\left\{x\in X\bigg\vert\lvert f_{n_k}(x)-f_{n_{k+1}}(x)\rvert\geq\frac{1}{2^k}\right\}

If k\leq i\leq j then for every x not in E_k\cup E_{k+1}\cup E_{k+2}\cup\dots we have

\displaystyle\lvert f_{n_i}(x)-f_{n_j}(x)\rvert\leq\sum\limits_{m=i}^\infty\lvert f_{n_m}(x)-f_{n_{m+1}}(x)\rvert<\frac{1}{2^{i-1}}

That is, the subsequence \{f_{n_k}\} is uniformly Cauchy on the set X\setminus\left(\cup_{i\geq k}E_i\right). But we also know that

\displaystyle\mu\left(\bigcup\limits_{m=k}^\infty E_m\right)\leq\sum\limits_{m=k}^\infty\mu(E_m)<\frac{1}{2^{k-1}}

and so \{f_{n_k}\} is almost uniformly Cauchy, as asserted.

Finally, we can take this subsequence that is almost uniformly Cauchy, and see that it must be a.e. Cauchy. We write f(x)=\lim_kf_{n_k}(x) at all x where this sequence converges. And then for every \epsilon>0,

\displaystyle\left\{x\in X\big\vert\lvert f_n(x)-f(x)\rvert\geq\epsilon\right\}\subseteq\left\{x\in X\bigg\vert\lvert f_n(x)-f_{n_k}(x)\rvert\geq\frac{\epsilon}{2}\right\}\cup\left\{x\in X\bigg\vert\lvert f_{n_k}(x)-f(x)\rvert\geq\frac{\epsilon}{2}\right\}

The measure of the first set on the right is small for sufficiently large n and n_k by the assumption that \{f_n\} is Cauchy in measure. The measure of the second approaches zero because almost uniform convergence implies convergence in measure.

And thus we conclude that if \{f_n\} is Cauchy in measure, then there is some measurable function f(x) to which \{f_n\} converges in measure. The topology of convergence in measure may not come from a norm, but it is still complete.


May 21, 2010 - Posted by | Analysis, Measure Theory


  1. […] sequences of integrable simple functions, then they’re both also Cauchy in measure, which implies that they each converge in measure to some function. If they converge to the same function (a.e.) , […]

    Pingback by Indefinite Integrals and Convergence II « The Unapologetic Mathematician | June 1, 2010 | Reply

  2. a counter example of a sequence of functions on the real interval E=[0,1] would be the sequence of
    indicator functions on the intervals [0,1/2] [1/2,1] [0,1/3] [1/3,2/3] [2/3,1] [0,1/4]…

    it is convergent in measure to the null function but not pointwise convergent almost everywhere.
    or am i wrong?

    Comment by tom | September 13, 2011 | Reply

  3. I’m sorry, there’s a problem somewhere in here, but this is an ancient post, I don’t even remember the reference I was using, and I have a real job I have to attend to; I can’t fix this at the moment.

    Comment by John Armstrong | September 13, 2011 | Reply

  4. Just for those who are (still) interested: Convergence in measure and a.e. convergence are not equivalent (even on sets of finite measure), as tom’s example shows. But the following holds:

    If E is of finite measure and f_n converges a.e. to f, then f_n converges to f in measure.

    Conversely, if f_n converges to f in measure, there exists a subsequence of f_n which converges to f a.e.

    The existence of a subsequence in the latter implication is the critical point here.

    Comment by Daniel | February 16, 2014 | Reply

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