# Is it correct to interpret the "dx" in the standard notation for integrals as the Lebesgue measure?

Ok so I am in my Calc I class for the summer and we are just beginning to talk about integrals. I know a little bit about measure theory and the Lebesgue integral and why is it more general than the Riemann integral and today my teacher said something that really irked me. She said that the "dx" is the same thing as the differential of a function that we talked about a chapter previously. This didn't sit right with me at all as I have always interpreted it as the Lebesgue measure. I know it can also be interpreted as a differential form but I know next to nothing about differential geometry, so let's just stick to the measure theoretical interpretations (though I have a feeling they are somehow related). So what I am asking is whether interpreting the "dx" as the Lebesgue measure formally correct?

• No. Every Riemann $\int$ble function is Lebesgue $\int$ble with the same value, but not conversely. Jun 27, 2016 at 4:36
• The symbol $dx$ in Riemann integral is just a formal placeholder for the dummy variable. And this is even true in the Lebesgue-measure-theory picture (unless you designate $x$ itself as the Lebesgue measure so that this fits into the notation $\int f \, d\mu$). It is the differential geometry picture that gives a meaningful, separate definition to the symbol $dx$ where it is defined as a 1-form. Jun 27, 2016 at 5:44

The "dx" is not the Lebesgue measure, no.

I took Riemann calculus long before I took Lebesgue calculus, so forgive me if I go too fast or too slow over any of the individual topics here -- I'm not sure which parts you've covered and which parts you haven't.

The idea behind a Riemann Sum is to find rectangles whose height match the function at given points, and the sum of the rectangles estimates the area under the curve of the function. If we create a finite number of rectangles this way, the result is often an approximation of the area, but as we make more and more rectangles with smaller and smaller widths, the error in the approximation gets smaller and smaller.

Compare this for a moment with Lebesgue measure; a rectangle has a 2-dimensional Lebesgue measure equal to its area, and a line has a 1-dimensional Lebesgue measure equal to its length. As we decrease the width of the rectangle, we have a sense that the rectangle is certainly getting smaller, but we don't really worry about the rectangle "going away" because we know that even the line still has a 1-dimensional measure.

Now, at this point, we're adding up rectangles that are getting smaller and smaller, but the area under the curve is converging faster than the width is shrinking, so we're left with the sum of these pairs of "height $\times$ tiny width", and yet we can't just leave out the width (or say zero), because that would misrepresent what we're adding. So instead we represent it with dx.

In the Lebesgue integrals, you sometimes see something like $\int_A f(x)d\mu$, where $\mu(S)$ represents the Lebesgue measure of S. You may even see a notation something like $d\mu = \mu(dx)$. The general point being made, however, is that the Lebesgue integral is integrating the function in relation to the Lebesgue measure in the same sense that our Riemann sum is integrating the function in relation to x, or the x-direction (the direction of infinitely thin slices).

While I can see where you might be tempted to think of this $dx$ as Lebesgue measure, it will be much more helpful for you to interpret the $dx$ as just representing the phrase "with respect to x" until you get a clearer feel for the differential and how it fits into the scheme of things.

Presumably all of the integrals you are seeing are Riemann integrals.

On any compact (read: finite) interval, the Riemann integral and the Lebesgue integral coincide for bounded functions.

However, the Riemann and Lebesgue integrals do not coincide for improper integrals (even for bounded functions).

So for every (Riemann) integral you would likely ever come across, as long as it is not an improper integral, therefore has finite bounds of integration, and the function is not unbounded over the domain of integration, then the integral is equal to the Lebesgue integral over the same interval.

Therefore in this sense it would be appropriate in most cases. My experience, however, was that many calculus classes don't emphasize the distinction between regular (Riemann) integrals and improper integrals very much, to the point that students think they are actually just the same, and that the limit used to evaluate the improper integral is just a formalism which makes it easier to calculate -- not so.

Again, in most normal cases, yes you could consider it as such. But when improper integrals are concerned, even with bounded functions, in general you cannot. For example: $\displaystyle\frac{\sin x}{x}$ is has a defined improper Riemann integral over the entire real line, but is not Lebesgue integrable (this is the standard counterexample with which I am familiar).

See also, for example: General condition that Riemann and Lebesgue integrals are the same

To add to Vaekor's Answer: The differential geometry picture of $\int f(x)dx$ is that $f(x)dx$ is what is known as a differential $1$-form (see: https://en.wikipedia.org/wiki/Differential_form). It is an object that lives in the cotangent space of the manifold (which is just $\mathbb{R}$ in this case). There are higher order forms, which do come up in multivariable calculus, for instance you might find in your book in later chapters, double integrals such as $\int\int f(x,y)dxdy$. This is more properly written as $\int\int f(x,y)dx\wedge dy$, where the symbol $\wedge$ denotes what's called the wedge product, this is all part of what is known as the exterior algebra, or Grassmann algebra (the first few pictures at https://en.wikipedia.org/wiki/Exterior_algebra show some of the important concepts of this construction). The usual antidifferentiation procedure for solving a definite integral invokes what is known as the interior product, and for multiple integrals, this process can (provided certain conditions are met) be carried out iteratively (the conditions are the hypothesis of the Generalized Stokes' Theorem, and the conclusion of that theorem is that you can in fact solve multiple integrals by iterative single integrals).