Integrable (summable) functions

[dimitri151]

Please look at attached pdf (I can't do latex in this editor).

 

[micromass]

Allright, a function f is defined to be integrable if it is measurable and both f^+ and f^- are integrable. Thus if you see the statement: "let f be integrable", then this will automatically imply that f is measurable and that f^+ and f^- be integrable.

So, in you words, we indeed have (f integrable) <=> (f measurable, f^+ and f^- integrable)

Furthermore, we have DEFINED that $$\int_E f=\int_E f^++\int_E f^-$$. So if you want to calculate $$\int_E f$$, then the only way to do it is by implying this property. You cannot calculate it in another way (unless you have a theorem to prove it).

Now, your counterexample. You have indeed shown that f^+ and f^- are infinite. This means by definition that f is not integrable and that $$\int_{\mathbb{R}^+}f$$ does not exist.
However, you have calculated the integral as

$$\lim_{n\rightarrow+\infty}{\int_{[0,n]}{f}}$$

While for Riemann integrals, this was OK (because it was essentially the definition of an extended Riemann integral). This is certainly not OK for Lebesgue integrals. This means that you can not calculate the Lebesgue integral this way. You have actually found a nice example of a function which can be calculated in the conventional sense, but which is not Lebesgue-integrable.

Note that the above is not entirely lost. In some cases we DO have that

$$\int_{\mathbb{R}^+}{f}=\lim_{n\rightarrow +\infty}{\int_{[0,n]}{f}}$$.

You will see when this holds in the so called monotone convergence theorem and dominated convergence theorem. However, the hypotheses of these theorems will fail in your example.

Another nice example is $$\int_0^{+\infty}{\frac{\sin(x)}{x}}$$. This is not Lebesgue-integrable (in the sense that both f^+ and f^- are not integrable). But we can calculate it as extended Riemann integrals. I.e. we can give meaning to

$$\int_0^{+\infty}{\frac{\sin(x)}{x}}=\lim_{x\rightarrow +\infty}{\int_0^x{\frac{\sin(x)}{x}}$$

but then the above integral is a Riemann integral and NOT a Lebesgue integral.

Finally, note that such a situations can not occur on compact intervals. In a compact interval, it holds that every Lebesgue integral is Riemann integral...

 

[pwsnafu]

Can you have f integrable over E and f +, f − not integrable over E?

The Lebesgue integral is a definite integrable, that means f is integrable iff |f| is integrable. For a function to be an element of L¹, we need
$$\int_E |f| \, d\mu < \infty$$

 

[dimitri151]

Thanks for the comprhensive reply, micromass. How come you can't use the dominated convergence theorem. If you let f_n={f, x<n; 0, x>=n} then f_n converges pointwise to f. Furthermore, f_n and f are dominated by g=1. So why then doesn't lim integral(f_n)=integral(f)?

Oh i got it. Because 1 isn't integrable over R+.

 

[Stephen Tashi]

dimitri151,

There is a bug in the way the forum's message editor displays latex. When composing a message with LaTex in it, you must do a "Preview" of the page and then reload the page in your browser. Before the page is reloaded, the latex my look crazy.

 

[dimitri151]

The source of my confusion was the way the definition was phrased. In that case the definition is of the form 'If such and such holds about x then we say x is so and so'. It looks superficially like an 'If A then B' statement, but it isn't. Like was pointed out, it's more like 'If A then B and if B then A' or in other words 'A if and only if B'. Thanks again for the clarifying remarks.

 

[micromass]

The source of my confusion was the way the definition was phrased. In that case the definition is of the form 'If such and such holds about x then we say x is so and so'. It looks superficially like an 'If A then B' statement, but it isn't. Like was pointed out, it's more like 'If A then B and if B then A' or in other words 'A if and only if B'. Thanks again for the clarifying remarks.

I know it can be confusing at first, but I've always seen definitions stated this way. But it's great that you've asked!