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.
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:
$$ \int_X f(x) \, d\mu(x) = \int_{0}^{\infty} \mu{\{x \in X: f(x) \geq t \}} \, dt $$
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 &quot;area under the measure&quot;. 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)