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E,

I am working on some homework problems in which we are being asked to evaluate integrals using measure integration.

An example problem I am working on is to evaluate:

$$ \int_{[0,1]} f\,dm$$ Where $f(x) = x$ for every $x \in \mathbb R$ and $m$ is Lebesgue Measure.

I am posting here because I am struggling to understand how exactly to evaluate a measure integral and what the answer should even look like.

Firstly, I would like to explain what I understand and what we have covered so far in lectures. So far in our latest class we proved The Dominated Convergence Theorem and have of course shown recently the key properties of Lebesgue integrals along with proving The Monotone Convergence Theorem and Fatou's lemma.

I understand that we often write out simple functions in their canonical representation and can evaluate integrals that way however I have never seen a worked example. I also understand the following definition presented in class:

Let $f:X\to [0,\infty]$ be measurable then define: $\int_X fd\mu := \sup\{\int_X \phi d\mu \mid\phi: X \to [0,\infty)$ is simple & measurable with $0 \le \phi \le f\}$

What I do not understand and is causing me much confusion is how to apply the definitions and theorems to actually solve an integral.

I would greatly appreciate an explanation or some suggestions on how to evaluate such an integral and how to better understand these types of exercises rather than just an answer if possible. I would like to fill in the gaps in my understanding that is preventing me from knowing how to answer these type of questions despite knowing and having some understanding of the course material.

Thanks!

P.S: Please do not hesitate to suggest any formatting or editing corrections I have no experience with MathJax.

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  • $\begingroup$ Do you know a nonnegative Riemann-integrable function is also Lebesgue integrable with the same values? This fact helps a lot in your question. $\endgroup$ – pisco Sep 3 '17 at 6:57
  • $\begingroup$ @pisco125 I think he knows it but is asking to do the integral from the definition of Lebesgue integral. Otherwise the exercise is trivial and doesn't give any insight. $\endgroup$ – Harto Saarinen Sep 3 '17 at 7:01
  • $\begingroup$ @pisco125 Thank you for the reply but we have not yet stated that in class. But that helps because that means the integral should be equal to a half. $\endgroup$ – Legendary Sep 3 '17 at 7:05
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    $\begingroup$ A few TeX suggestions: Put \, between $f$ and $d\mu$ to make a little space. Use \sup instead of Sup to get $\sup$ instead of $Sup$. Use \mid to get a separator with some space around it in a set constructor: $\sup \{ \int_X \phi \, d\mu \mid \phi : X \to [0, \infty) \}$. $\endgroup$ – md2perpe Sep 3 '17 at 7:21
  • $\begingroup$ @md2perpe Thank you! I have just added these corrections! $\endgroup$ – Legendary Sep 3 '17 at 7:25
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Since you know dominant convergence theorem, we can make use of it.

Define, for $i=1,2,..,n$, $$f_n(x) = \frac{i}{n} \quad \text{when } x\in [\frac{i-1}{n},\frac{i}{n})$$ Then $x\leq f_n(x)$ for all $x\in [0,1)$. Note that $f_n(x) $ converge pointwise to $x$. Dominant convergence theorem says: $$\int_{[0,1)} x dx = \lim_{n\to\infty}\int_{[0,1)} f_n(x) dx$$ $f_n(x)$ is a step function, which makes the limit easy to calculate. We have $$\int_{[0,1]} f_n(x) dx = \frac{1}{n}\left( \frac{1}{n} + \frac{2}{n} +... + \frac{n}{n} \right) = \frac{1}{n}\frac{n(n+1)}{2n}$$

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  • $\begingroup$ Thank you for the answer, but could I please ask you the following questions. We have changed the domain of the integral from [0,1] to [0,1), why is this allowed? And when applying the DCT we do not need to consider that the dx's in the integrals are actually the Lebesgue measure? I can follow most of this but still confused. $\endgroup$ – Legendary Sep 3 '17 at 7:23
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    $\begingroup$ Sorry for confusion. $dx$ here all means Lebesgue integration, they have nothing to do with Riemann integral. Sometimes people write this way as long as the context is clear. You may replace $dx$ with $dm$ if you feel uncomfortable. $\endgroup$ – pisco Sep 3 '17 at 7:25
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    $\begingroup$ You can define $f_n(1) = 1$, then dominate convergence theorem still applies and the answer is just the same. In fact, we often do not care about a set of measure zero (in the case, the point $1$) when doing Lebesgue integration. $\endgroup$ – pisco Sep 3 '17 at 7:27
  • $\begingroup$ That clears up my confusion thank you, and I can now understand all the steps here as well as the reasoning. If you do not mind one last question when we evaluate the integral the limit as n goes to infinity of f_n(x) should be 1? However we agreed earlier the integral should be the same as if it was done via Riemann Integration so how do we get the factor of 1/2? $\endgroup$ – Legendary Sep 3 '17 at 7:40
  • $\begingroup$ I added this detail to my answer. $\endgroup$ – pisco Sep 3 '17 at 7:44

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