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I have just read the Wikipedia article on the Lebesgue Integral, as well as this answer on Lebesgue integral basics.

I'm not sure how to interpret the differential $d \mu(x) = d \mu$ in the case where $\mu$ is the Lebesgue measure on $\mathbb{R}$.

For instance in a simple example, say

$$\int_{a}^{b} x d \mu (x)$$

For the Riemann integral $$\int_{a}^{b} x dx$$ the interpretation of $dx$ is an infinitesimal change in $x$, summing up the signed areas of the rectangles formed by sampling heights from points $f(x^*)$ in the subintervals as a limit.

For the Lebesgue integral, you partition the range of $f$ into infinitely fine intervals $[y_i, y_{i+1}]$ with $y_0=a, y_N=b$ and compute $\lim_{n \rightarrow \infty} \sum_{i=1}^{N} \mu (E_i) y_i^*$ where $E_i$ is the pre-image of the i-th partition interval, a subset of $R$ (the $x$-values), and $y_i*$ is a sample point within the $y$ range. My confusion seems to be that $\mu$ is a measure of sets of $y$ values, but it's written as a function of $x$ in the integral notation? Any insights appreciated.

Edit: as per the comments, $\mu$ is a measure of measurable sets of $x$ values which are in the pre-image of some set of y-values, but still the point remains that $d \mu(x)$ seems to input individual $x$ values as per the notation.

IntegrateThis
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    The "small thing" is the measure of a small set on which you approximate the integrand as being constant. Note that what you measure is indeed back in the domain, not the codomain; it is the measure of the preimage and the preimage pulls you back into the domain. So you are still taking "size in domain times value in codomain" just like in Riemann integration, the geometric difference is that you draw horizontal lines to divide up the area instead of vertical lines. – Ian Nov 02 '21 at 22:27
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    You are possibly reading too much into the notation. There are 3 objects of interest, the function you are integrating, the measure and the set over which you are integrating. The measure 'operates' on subsets of the domain, not the codomain. To a very simplistic degree, the point of Lebesgue integration is that it deals with finer subsets of the domain than rectangles as is the case with Riemann. – copper.hat Nov 02 '21 at 22:34
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    @copper.hat, right so I made an error. If I understand correctly, then the measure operates on subsets of the domain, where I wrote above that $\mu$ inputs $y$ values, rather the $\mu(E_i)$ is the measure of the pre-image of a range of y-values. – IntegrateThis Nov 02 '21 at 22:37
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    i always preferred the notation $\int f(x), \mu(\mathrm{d}x)$, as it is clear and it is intuitive (to me), but this does not seem to be as popular. – Brian Shin Nov 02 '21 at 22:45

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