# Let $f$ integrable over $E$ and $C$ a measurable subset of $E.$ Show that $\int_{C} f = \int_{E} f. \chi_{C}.$

I am trying to prove the following result for characteristic functions first and then simple functions:

"Let $$f$$ integrable over $$E$$ and $$C$$ a measurable subset of $$E.$$ Show that $$\int_{C} f = \int_{E} f. \chi_{C}$$"

Here is a trial in case of Characteristic functions:

Define $$\chi_C(x) = \begin{cases} 1 & \text{if } x \in C \\ 0 & \text{if } x \in E \sim C. \end{cases}.$$

Then, $$\int_E f. \chi_C = \int_C f. \chi_C + \int_{E \sim C} f. \chi_C = \int_C f. 1 + \int_{E \sim C} f. 0 = \int_C f$$

My question is: Is my proof correct?

Here is a trial for simple functions:

For each $$i$$ where $$1 \leq i \leq n,$$ let $$a_i$$ be the distinct values taken by $$\psi$$ and $$E_i = \psi^{-1}(a_i) = \{x \in E | \psi (x) = a_i \}$$ are disjoint, Thus $$\psi = \sum_{i=1}^{n} a_i \chi_{E_i} = \sum_{i=1}^{n} a_i \chi_{E_i \cap C} + \sum_{i=1}^{n} a_i \chi_{E_i \cap C^c}.$$ But then $$\int_E \psi = \sum_{i=1}^{n} a_i . m(E_i \cap C) + \sum_{i=1}^{n} a_i . m(E_i \cap C^c).$$

My question: I am confused how the integration of $$\psi$$ will look like if $$a_i's$$ took negative values? Also, How will I go to the case of bounded function defined on a set of finite measure ?Could anyone clarify this to me please?

Yes, your proof for characteristic functions case is fine.

I guess that you're trying to use the standard machine to extend that result to all bounded measurable functions. It usually goes like this :

Step 1 : Prove the property for all characteristic functions.
Step 2 : Using linearity, extend the property to all simple non-negative functions
Step 3 : Using the Monotone Convergence Theorem extend the property to all bounded measurable non-negative functions
Step 4 : Extend the property in question to all bounded measurable functions $$h$$ by writing $$h=h^+ - h^−$$ and using linearity of Lebesgue integral.

You are at step 2, you thus only need to prove the result for non-negative simple functions first, so take all your $$a_i$$'s to be non-negative.

• why I am extending the property to all simple non-negative functions only ?

As I said, this is just step 2 out of 4 in the standard machine, which will then allow you to extend the property to all measurable functions. If you're not familiar with this approach, it is a very effective and thus very popular way to generalize results which are true for simple functions to general bounded measurable functions.
The idea is that, instead of trying to prove directly a result for all measurable functions, you first prove it for characteristic functions, which are generally much easier to deal with, and from there progressively generalizing to non-negative simple functions, non-negative measurable functions and finally arbitrary measurable functions. You have examples of this method being used to solve a some measure theory problems e.g. here, here or here.
Lastly, here is a quote from Chapter 3 of Prof. Gordan Žitković's lecture notes on probability theory, which essentially restates what I just wrote (some more examples are given in the notes) :

The Lebesgue integral is very easy to define for non-negative simple functions and this definition allows for further generalizations. In fact, the progression of events you will see in this section is typical for measure theory: you start with indicator functions, move on to non-negative simple functions, then to general non-negative measurable functions, and finally to (not-necessarily-nonnegative) measurable functions. This approach is so common, that it has a name - the Standard Machine

I suggest you read further on the topic to familiarize yourself with the idea.

• How to prove the property for all non-negative simple functions ?

As you wrote, if $$\psi$$ is a non-negative simple function, it can be written as $$\psi = \sum_{i=1}^{n} a_i \chi_{E_i}$$. And similarly $$\psi\cdot\chi_{C} = \sum_{i=1}^{n} a_i \chi_{E_i}\cdot\chi_{C}$$ So using the linearity of Lebesgue integral and the fact that you have proven the property for characteristic functions \begin{align} \int_E \psi\cdot\chi_C &= \int_E \sum_{i=1}^{n} a_i \chi_{E_i}\cdot\chi_{C}\\ &=\sum_{i=1}^{n} a_i \int_E \chi_{E_i}\cdot\chi_C\tag1\\ &= \sum_{i=1}^{n} a_i \int_C \chi_{E_i}\tag2\\ &= \int_C \sum_{i=1}^{n} a_i \chi_{E_i} = \int_C \psi \tag3\\ \end{align} Where I used the linearity of Lebesgue integral in $$(1)$$ and $$(3)$$, and the fact that the property is true for characteristic functions (which you have proven) in $$(2)$$.

• should not $f$ be a characteristic function also in my proof of the statement in case of characteristic functions?
– user965463
Commented Nov 16, 2021 at 11:54
• Also, I do not know how can I complete the proof for non-negative simple function. Any help in this please?
– user965463
Commented Nov 16, 2021 at 12:00
• Also, why I am extending the property to all simple non-negative functions only?
– user965463
Commented Nov 16, 2021 at 12:02
• How this statement $$\int_{C} f = \int_{E} f. \chi_{C}$$should look like in case of simple functions? what is $\psi . \chi_{C}$?
– user965463
Commented Nov 16, 2021 at 12:09
• I edited my answer to address your questions, hope that helps ! Oh and of course in your proof of the statement for characteristic functions, you have to assume that $f$ is a characteristic function, otherwise it's kind of pointless.... I thought that was what you did. Commented Nov 16, 2021 at 13:30