The simplest example of a non-Turing-recognizable language is the complement of $A_{\text{TM}}$, $$\overline{A_\text{TM}}=\{\langle M,w\rangle\mid M\text{ is a TM and }M\text{ does not accepts }w\}.$$

Let us reduce $\overline{A_\text{TM}}$ to $L$.

Given a TM $M$ and a string $w$, construct the following TM.

$M'$= "On input $s$:

1. Simulate $M$ on input $w$ for $|s|$ steps using less than $|s|^3+159759$ steps.
2. If the simulated $M$ has halted, possibly before $|s|$ steps, run forever. Else halt."

Here $3$ and $159759$ just mean some large-enough numbers independent of $M$ so that the simulation for $|s|$ steps can be done.

Verify that $M\in A_{\text{TM}}$ iff $M'\in L$.

Since $A_{\text{TM}}$ is not Turing recognizable, neither is $L$.

The simplest example of a non-Turing-recognizable language is the complement of $A_{\text{TM}}$, $$\overline{A_\text{TM}}=\{\langle M,w\rangle\mid M\text{ is a TM and }M\text{ does not accepts }w\}.$$

Let us reduce $\overline{A_\text{TM}}$ to $L$.

Given a TM $M$ and a string $w$, construct the following TM.

$M'$= "On input $s$:

1. Simulate $M$ on input $w$ for $|s|$ steps using less than $|s|^3+159759$ steps.
2. If the simulated $M$ has accepted, possibly before $|s|$ steps, run forever. Else halt."

Here $3$ and $159759$ just mean some large-enough numbers independent of $M$ so that the simulation for $|s|$ steps can be done.

Let us verify that $\langle M,w\rangle\in \overline{A_\text{TM}}$ iff $\langle M'\rangle\in L$.

• If $\langle M,w\rangle\in \overline{A_\text{TM}}$, i.e., $M$ does not accept $w$, then on input $s$, $M'$ will always halt in $|s|^3+159759$ steps. Hence $\langle M'\rangle\in L$.
• If $\langle M,w\rangle\notin \overline{A_\text{TM}}$, i.e., $M$ accepts $w$, then for input $s$ that is long enough, the simulated $M$ will have accepted after step 1. Hence $M'$ will run forever on those input. Hence $\langle M'\rangle\notin L$.

Since $\overline{A_{\text{TM}}}$ is not Turing recognizable, neither is $L$.