Consider a measurable space $(S,\Sigma)$ and a finite non-atomic measure $\mu$. If $f:S\rightarrow \mathbb{R}$ is integrable and non negative (and $f\neq 0$), then let $\lambda$ be the measure on $(S,\Sigma)$ defined by $$\lambda(E)=\int_E fd\mu\quad \text{for every }E\in\Sigma$$ My question is, is $\lambda$ non atomic as well? If $f=\sum_{i} a_i\chi_{A_i}$ is simple and non-negative then this should be true, since we get $$\lambda(E)=\sum_{i} a_i\mu(A_i\cap E)$$ Now define the measures $\mu_i(E)=a_i\mu(A_i\cap E)$. Since each $\mu_i$ is also non atomic, it follows that $\lambda$ is the sum of non-atomic measures, which is known to be non-atomic as well.
However, it is not clear to me how to generalize this to general non-negative integrable functions.
Suppose that $A$ is an atom for $\lambda$, and consider $\{ f > 0 \} \cap A = A'$. Then $A'$ is also an atom for $\lambda$. Consider $E \subset A'$ so that $0 < \mu(E) < \mu(A')$ , and note that $f$ is $0$ $\mu$-a.s. on either $E$ or $A' \setminus E$ (otherwise $\lambda(E) > 0$ and $\lambda( A' \setminus E) > 0$, which contradicts $A'$ atomic). However, this contradicts the assumption that $f$ is positive on $A'$.
Here is how I arrived at this argument (which is a train of thought that uses a hammer and also is a little informal at the end):
Suppose that there was an $A$ so that $\lambda(A) > 0$ but for all $E \subset A$ with $\lambda(E) < \lambda(A)$, then $\lambda(E) = 0$.
Let $E \subset A$ be such that $\mu(E) = (1/2) \mu(A)$ (such an $E$ exists by Sierpinski's theorem: https://en.wikipedia.org/wiki/Atom_(measure_theory)).
Then $f$ is a.e. $0$ on either $E$ or $A \setminus E$. Otherwise, $\lambda(E) > 0$ and $\lambda(A \setminus E) > 0$, which implies that $A$ wasn't an atom for $\lambda$.
Without loss of generality, suppose that $f = 0$ a.s. on $E$.
Now repeat infinitely many times, to get that $f = 0$ a.s. on $A$, which is a contradiction.