What is the name of this theorem, and are there any caveats?

A probability density function is usually defined in the following way:

Let $X$ be a random variable. Then $f$ is the probability density function of the distribution given by $X$ if and only if $f(x)\geq 0$ for all $x\in\mathbb{R}$ and $$ P(X\in A)=\int_A f(x)\,\lambda(\mathrm d x), \quad A\in\mathcal{B}(\mathbb{R}). \tag{1} $$

However, $(1)$ is equivalent to both of the following

$$ P(a\leq X\leq b)=\int_a^bf(x)\,\lambda(\mathrm dx),\quad -\infty<a<b<\infty.\tag{2} $$

and

$$ \mathrm{E}[\varphi(X)]=\int_\mathbb{R}\varphi(x)f(x)\,\lambda(\mathrm dx),\tag{3} $$ for every measurable and integrable (with respect to the measure $f\lambda$) function $\varphi$.

It is clear that $(3) \Rightarrow (2)$ (take $\varphi=1_A$) and $(1)\Rightarrow (2)$ (take $A=[a,b]$). But on the other hand $(2)\Rightarrow (1)$ by the use of Dynkin's lemma and also $(1)\Rightarrow (3)$ by a standard argument that is often used in probability theory. The argument goes as follows:

a) The property $(3)$ holds for all indicator functions according to $(1)$.

b) If $\varphi$ and $\psi$ are two functions satisfying $(3)$, then $$ \begin{align*} \mathrm{E}[(\varphi+\psi)(X)]=\mathrm{E}[\varphi(X)]+\mathrm{E}[\psi(X)]&=\int_\mathbb{R}\varphi(x)f(x)\,\lambda(\mathrm dx)+\int_\mathbb{R}\psi(x)f(x)\,\lambda(\mathrm dx)\\ &=\int_\mathbb{R}(\varphi+\psi)(x)f(x)\,\lambda(\mathrm dx), \end{align*} $$ and so $\varphi+\psi$ satisfies $(3)$. In a similar fashion one can show that $\mathrm{E}[\alpha \varphi(X)]=\alpha\mathrm{E}[\varphi(X)]$ for $\alpha\in\mathbb{R}$ and hence the set of functions satisfying $(3)$ form a vector space.

c) Suppose that $(\varphi_n)_{n\geq 1}$ is a sequence of non-negative, increasing functions satisfying $(3)$ such that $\varphi=\lim_{n\to\infty}\varphi_n$ exists pointwise. Then by applying the monotone convergence theorem (twice) we have $$ \mathrm{E}[\varphi(X)]=\lim_{n\to\infty}\mathrm{E}[\varphi_n(X)]=\lim_{n\to\infty}\int_\mathbb{R}\varphi_n(x)f(x)\,\lambda(\mathrm dx)=\int_\mathbb{R}\varphi(x)f(x)\,\lambda(\mathrm dx). $$

Now the standard argument yields that $(3)$ holds for every measurable and integrable (with respect to the measure $f\lambda$) $\varphi$.

This also shows that $\varphi$ in $(3)$ could just as well have been chosen to be "positive and bounded".