# Is Lebesgue integral area under the curve?

We know that Riemann integral is defined to be the area between the graph of a function $y=f(x)$ and $x$-axis. Actually, many people state the definition works in both ways (area under the curve is defined to be equal to Riemann integral).

If a Riemann integral of $f$ exists, then Lebesgue integral also exists and is equal to Riemann integral. However, there are Lebesgue-integrable functions that are not Riemann-integrable.

Does Lebesgue integral also give the area under the curve of such functions? Can it be proven formally? Actually this is an important question, because it depends on whether the area under a curve is defined as the value of integral or not.

• (assuming $f \geq 0$) the Riemann integral is the area Sep 27 '15 at 17:34
• I think if a function is not Riemann integrable, then its graph is not a curve. Also, their graphs are very complicated so that it may be hard to talk about the area under it. Sep 27 '15 at 18:01
• The Riemann integral is defined as a limit of sums. It's not defined as the area of anything. Rather the area under a curve, if it's defined at all, is defined to be the Riemann integral.
– zhw.
Sep 27 '15 at 18:15

I would say the answer to your question is yes, because if the function is non-negative, you can write the Lebesgue integral as a Riemann integral: $$\int_X f(x) \, d\mu(x) = \int_{0}^{\infty} \mu{\{x \in X: f(x) \geq t \}} \, dt$$

The Lebesgue integral is an extension to the integral definition so as to give sensible results even for functions that aren't Riemann integrable. If the function is Riemann integrable (i.e., the "area under the curve" makes sense) both agree.

• Yes, but my question is about functions that aren't Riemann integrable. Does Lebesgue integral equal the area under the curve for such functions? Sep 27 '15 at 17:39
• @user216094 Define area in such cases. .. Sep 27 '15 at 17:59

This question is a bit philosophical: it amounts to asking "is the Lebesgue integral a good way of defining area?". One thing that might clarify the situation is that if $f$ is a nonnegative measurable function on $[a,b]$, then

$$\int_a^b f(x) dx = \int_a^b \int_0^{f(x)} dy dx.$$

That is, the Lebesgue integral is the two-dimensional Lebesgue measure of the region between the curve and the $x$-axis. So if the two-dimensional Lebesgue measure is how you define area, then the answer to your question is yes.

• Yes, it's quite philosophical. Look at Weierstrass function - it's Riemann integrable, but how can you think of the area of a function if you can't even see its graph? Sep 27 '15 at 18:30

I was searching for an answer to that same question (which led me to this page, albeit 6 years too late). Since I didn't find anything completely satisfying I made my own attempt at an answer with an example where to me the concept of area isn't clear. Consider $$f: X \rightarrow Y$$ with domain $$X = \{a,b,c\}$$ and $$Y=\{ 0,1\}$$ (this happens to be an indicator function). And $$f = \{(a,0),(b,0),(c,1) \} \subset X \times Y$$. If we use counting measure (cardinality) the Lebesgue integral $$\int_X f d \mu = 0 \cdot \mu(f^{-1}(\{0\}) + 1 \cdot \mu(f^{-1}(\{1\}) = 0 \cdot |\{a,b\}| + 1 \cdot |\{c\}| = 1$$ Here, $$f$$ is a discrete function so is there really a curve or an area under it? It seems to me the answer by Chappers produces "area under the measure". Does that translate to area under $$f$$?

It seems to me since there is nothing connecting the discrete points of $$f$$ this wouldn't be an area but this is still a question for me too so if this logic is incorrect, I would like to know why (maybe leave a comment so I can update it)